MH1810 Math

Chapter 6b: Integration

Imron Rosyadi

6.4 Techniques of Integration

So far, we’ve focused on basic integration rules and the Fundamental Theorem. Now we move into more advanced techniques for finding antiderivatives.

6.4.1 The Substitution Rule

The Substitution Rule is a direct consequence of the Chain Rule for differentiation. It helps simplify integrals by changing the variable of integration.

Important

Theorem 6.4.1 (Substitution Rule for Indefinite Integrals): \[ \int f(u(x))\underbrace{u'(x)dx}_{du} = \int f(u)du. \]

The idea behind the substitution rule is to replace a relatively complicated integral by a simpler integral.

Technique: Choose \(u\) to be some inner function in the integrand whose derivative (or a constant multiple of its derivative) also occurs in the integrand.

Think of it as reversing the chain rule. When you differentiate \(F(u(x))\), you get \(F'(u(x))u'(x)\). The substitution rule helps you go back from \(F'(u(x))u'(x)\) to \(F(u(x))\). The choice of ‘u’ is the most critical step.

Substitution Rule: Examples

Let’s practice applying the substitution rule.

Example 6.4.2: Evaluate \(\int \frac{x}{x^2 + 1} dx\).

Solution:

Notice that the derivative of \(x^2 + 1\) is \(2x\). The numerator has \(x\).

Let \(u = x^2 + 1\).

Then \(\frac{du}{dx} = 2x\), which means \(du = 2x \, dx\).

We have \(x \, dx\) in the integral, so we can write \(x \, dx = \frac{1}{2} du\).

Substitute \(u\) and \(du\) into the integral: \[ \int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{1}{2} \int \frac{1}{u} du \] Now integrate with respect to \(u\):

\(= \frac{1}{2} \ln |u| + C\)

Finally, substitute back \(u = x^2 + 1\):

\(= \frac{1}{2} \ln |x^2 + 1| + C\).

Since \(x^2 + 1\) is always positive, we can write it as \(\frac{1}{2} \ln (x^2 + 1) + C\).

The key here was recognizing that \(x\) in the numerator was part of the derivative of \(x^2+1\) in the denominator. This setup is a classic sign for u-substitution.

Substitution Rule: More Examples

Let’s try a few more.

Example 6.4.3: Evaluate \(\int \sin^3 x\cos xdx\).

Solution:

Notice that \(\frac{d}{dx} (\sin x) = \cos x\).

Let \(u = \sin x\).

Then \(du = \cos x \, dx\).

Substitute \(u\) and \(du\): \[ \int u^3 du \] Integrate with respect to \(u\):

\(= \frac{u^4}{4} + C\)

Substitute back \(u = \sin x\):

\(= \frac{\sin^4 x}{4} + C\).

Example 6.4.4: Evaluate \(\int \frac{e^{3x}}{\sqrt{1 - e^{6x}}} dx\).

Solution:

Note that \(e^{6x} = (e^{3x})^2\).

Also, \(\frac{d}{dx} (e^{3x}) = 3e^{3x}\).

Let \(u = e^{3x}\).

Then \(du = 3e^{3x} \, dx\), so \(e^{3x} \, dx = \frac{1}{3} du\).

Substitute \(u\) and \(du\): \[ \int \frac{1}{\sqrt{1 - u^2}} \left(\frac{1}{3} du\right) = \frac{1}{3} \int \frac{1}{\sqrt{1 - u^2}} du \] Recall that \(\int \frac{1}{\sqrt{1 - u^2}} du = \sin^{-1} u + C\).

\(= \frac{1}{3} \sin^{-1} u + C\)

Substitute back \(u = e^{3x}\):

\(= \frac{1}{3} \sin^{-1} (e^{3x}) + C\).

Example 6.4.3 is straightforward. Example 6.4.4 requires a bit more algebraic foresight to see that \(e^{6x}\) can be written as \((e^{3x})^2\), making \(e^{3x}\) a good candidate for substitution. Recognize the form of the \(\sin^{-1}\) derivative!

Substitution Rule for Definite Integrals

When using substitution with definite integrals, you must change the limits of integration.

Important

Theorem 6.4.5 (Substitution Rule for Definite Integrals): \[ \int_{a}^{b}f(u(x))u^{\prime}(x)dx = \int_{u(a)}^{u(b)}f(u)du. \]

This means you can change the variable and the limits of integration once, and then evaluate the new integral directly, without substituting back to the original variable.

This is a very important point! If you’re doing a definite integral, changing the limits of integration from x-values to u-values saves you a step, as you don’t have to convert back to x before evaluating. Make sure to change both limits!

Substitution Rule for Definite Integrals: Example

Let’s illustrate the process with an example.

Example 6.4.6: Evaluate \(\int_0^8\frac{\cos\sqrt{x + 1}}{\sqrt{x + 1}} dx\).

Solution:

1. Choose \(u\) and find \(du\): Let \(u = \sqrt{x + 1} = (x+1)^{1/2}\). Then \(\frac{du}{dx} = \frac{1}{2}(x+1)^{-1/2} = \frac{1}{2\sqrt{x+1}}\). So, \(du = \frac{1}{2\sqrt{x+1}} \, dx\). This means \(\frac{1}{\sqrt{x+1}} dx = 2 \, du\).

2. Change the limits of integration:

   • When \(x = 0\), \(u = \sqrt{0 + 1} = \sqrt{1} = 1\).
   • When \(x = 8\), \(u = \sqrt{8 + 1} = \sqrt{9} = 3\).

3. Substitute into the integral and evaluate: \[ \int_ {0} ^ {8} \frac {\cos \sqrt {x + 1}}{\sqrt {x + 1}} d x = \int_ {1} ^ {3} \cos u \cdot (2 du) = 2 \int_ {1} ^ {3} \cos u \, du \] \[ = 2 \left[ \sin u \right]_1^3 = 2 (\sin 3 - \sin 1). \] (Note: \(\sin 3\) and \(\sin 1\) are in radians).

Notice how the limits of integration changed from 0 and 8 (for x) to 1 and 3 (for u). This is crucial for definite integrals. If you forget to change the limits, you must substitute u back to x before evaluating.

6.4.2 Integration by Parts

Integration by Parts is a technique derived from the Product Rule for differentiation. It’s used to integrate products of functions.

Important

Theorem 6.4.7 (Integration by Parts Formula): \[ \int u (x) \underbrace {v ^ {\prime} (x) d x} _ {d v} = u (x) v (x) - \int v (x) \underbrace {u ^ {\prime} (x) d x} _ {d u}. \] In short: \[ \int u \, dv = uv - \int v \, du \]

Note: The integrand is a product of 2 functions: one of which we choose to be \(u(x)\) and the other to be \(v'(x)\). Usually, we choose the function whose derivative simplifies as \(u(x)\) and the function we can easily integrate as \(v'(x)\). A common mnemonic for choosing \(u\) is LIATE:

• Logarithmic functions
• Inverse trigonometric functions
• Algebraic functions (polynomials)
• Trigonometric functions
• Exponential functions

Choose \(u\) as the function that comes first in the LIATE order.

Integration by parts is like the “product rule” for integration. The key challenge is strategically choosing which part of the integrand is ‘u’ and which is ‘dv’. The LIATE rule is a very helpful heuristic for making this choice, as it often leads to a simpler integral on the right side.

Integration by Parts: Examples

Let’s apply the integration by parts formula.

Example 6.4.8: Evaluate \(\int x\cos xdx\).

Solution:

Using LIATE, \(x\) is Algebraic, \(\cos x\) is Trigonometric. So, choose \(u=x\).

Let \(u = x \quad \Rightarrow \quad du = dx\)

Let \(dv = \cos x \, dx \quad \Rightarrow \quad v = \int \cos x \, dx = \sin x\)

(We don’t need the constant \(C\) for \(v\) here, as it would cancel out).

Now apply the formula \(\int u \, dv = uv - \int v \, du\): \[ \int x\cos xdx = x\sin x - \int \sin x \, dx \] \[ = x\sin x - (-\cos x) + C \] \[ = x\sin x + \cos x + C. \]

Integration by Parts: Examples

Example 6.4.9: Evaluate \(\int x^2\ln x dx\).

Solution:

Using LIATE, \(\ln x\) is Logarithmic, \(x^2\) is Algebraic. So, choose \(u=\ln x\).

Let \(u = \ln x \quad \Rightarrow \quad du = \frac{1}{x} dx\)

Let \(dv = x^2 dx \quad \Rightarrow \quad v = \int x^2 dx = \frac{x^3}{3}\)

Apply the formula: \[ \int x^2\ln x dx = (\ln x)\left(\frac{x^3}{3}\right) - \int \left(\frac{x^3}{3}\right)\left(\frac{1}{x}\right) dx \] \[ = \frac{x^3}{3}\ln x - \int \frac{x^2}{3} dx \] \[ = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2 dx \] \[ = \frac{x^3}{3}\ln x - \frac{1}{3}\left(\frac{x^3}{3}\right) + C \] \[ = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C. \]

Example 6.4.8 is a classic case where choosing u as the polynomial term simplifies things. Example 6.4.9 shows that sometimes the integral in the v du part is still a bit complex but simpler than the original, leading to a successful integration. Remember LIATE!

Integration by Parts: More Examples

Let’s continue with more diverse examples.

Example 6.4.10: Evaluate \(\int (t + 1)e^{t}dt\).

Solution:

Using LIATE, \(t+1\) is Algebraic, \(e^t\) is Exponential. So, choose \(u=t+1\).

Let \(u = t+1 \quad \Rightarrow \quad du = dt\)

Let \(dv = e^t dt \quad \Rightarrow \quad v = \int e^t dt = e^t\)

Apply the formula: \[ \int (t + 1)e^{t}dt = (t+1)e^t - \int e^t dt \] \[ = (t+1)e^t - e^t + C \] \[ = t e^t + e^t - e^t + C \] \[ = t e^t + C. \]

Example 6.4.11: Evaluate \(\int \tan^{-1}x dx\).

Solution:

This doesn’t look like a product, but we can treat it as \(\tan^{-1}x \cdot 1\).

Using LIATE, \(\tan^{-1}x\) is Inverse Trigonometric, \(1\) is Algebraic. So, choose \(u=\tan^{-1}x\).

Let \(u = \tan^{-1}x \quad \Rightarrow \quad du = \frac{1}{1+x^2} dx\)

Let \(dv = 1 \, dx \quad \Rightarrow \quad v = \int 1 \, dx = x\)

Apply the formula:

\[ \int \tan^{-1}x dx = x \tan^{-1}x - \int x \cdot \frac{1}{1+x^2} dx \] The new integral \(\int \frac{x}{1+x^2} dx\) can be solved using u-substitution (let \(w = 1+x^2\), \(dw = 2x dx\)).

\(\int \frac{x}{1+x^2} dx = \frac{1}{2}\int \frac{1}{w} dw = \frac{1}{2}\ln|w| + C_1 = \frac{1}{2}\ln(1+x^2) + C_1\).

So, \[ \int \tan^{-1}x dx = x \tan^{-1}x - \frac{1}{2}\ln(1+x^2) + C. \]

Example 6.4.10 is another good example of LIATE. Example 6.4.11 is a common trick for integrating inverse trigonometric functions: treat “1” as the dv part. It often requires a second integration technique (like u-substitution) to solve the \(\int v du\) part.

Reduction Formula

Sometimes, integration by parts needs to be applied multiple times. A reduction formula expresses an integral in terms of a simpler integral of the same form.

Example 6.4.12: Let \(I_{n} = \int x^{n}e^{x}dx\), where \(n\) is a non-negative integer. Prove that for \(n \geq 1\),

\[ I _ {n} = x ^ {n} e ^ {x} - n I _ {n - 1}. \]

Solution:

For \(n \geq 1\), we use integration by parts. Using LIATE, \(x^n\) is Algebraic, \(e^x\) is Exponential. So, choose \(u=x^n\).

Let \(u(x) = x^n \quad \Rightarrow \quad u'(x) = nx^{n-1} dx\)

Let \(v'(x) = e^x dx \quad \Rightarrow \quad v(x) = e^x\)

Apply the formula \(\int u \, dv = uv - \int v \, du\):

\[ I _ {n} = \int x ^ {n} e ^ {x} d x = x ^ {n} e ^ {x} - \int e ^ {x} (nx^{n-1}) d x \] \[ = x ^ {n} e ^ {x} - n \int x ^ {n - 1} e ^ {x} d x \]

Notice that \(\int x^{n-1}e^x dx\) is exactly \(I_{n-1}\).

So, we obtain the reduction formula:

\[ I _ {n} = x ^ {n} e ^ {x} - n I _ {n - 1}. \]

Reduction formulas are incredibly useful when you have integrals involving powers, where applying integration by parts once reduces the power by one. This allows you to iteratively solve the integral.

Using Reduction Formulas

Once a reduction formula is established, we can use it to find integrals for specific values of \(n\).

Example 6.4.13: Let \(I_{n} = \int x^{n}e^{x}dx\), where \(n\geq 0\). Use the reduction formula \[ I _ {n} = x ^ {n} e ^ {x} - n I _ {n - 1}, \] to determine a formula for \(I_4\).

Solution:

Apply the reduction formula repeatedly:

\(I _ {4} = x ^ {4} e ^ {x} - 4 I _ {3}\)

\(I _ {3} = x ^ {3} e ^ {x} - 3 I _ {2}\)

\(I _ {2} = x ^ {2} e ^ {x} - 2 I _ {1}\)

\(I _ {1} = x e ^ {x} - I _ {0}\)

We need to evaluate \(I_0\):

\(I_0 = \int x^0 e^x dx = \int e^x dx = e^x + C\).

Now substitute back up the chain:

\(I _ {1} = x e ^ {x} - (e ^ {x} + C) = x e ^ {x} - e ^ {x} - C\).

\(I _ {2} = x ^ {2} e ^ {x} - 2 I _ {1} = x ^ {2} e ^ {x} - 2 (x e ^ {x} - e ^ {x} - C)\) \(= x ^ {2} e ^ {x} - 2 x e ^ {x} + 2 e ^ {x} + 2 C\).

\(I _ {3} = x ^ {3} e ^ {x} - 3 I _ {2} = x ^ {3} e ^ {x} - 3 (x ^ {2} e ^ {x} - 2 x e ^ {x} + 2 e ^ {x} + 2 C)\) \(= x ^ {3} e ^ {x} - 3 x ^ {2} e ^ {x} + 6 x e ^ {x} - 6 e ^ {x} - 6 C\).

Finally, for \(I_4\): \(I _ {4} = x ^ {4} e ^ {x} - 4 I _ {3} = x ^ {4} e ^ {x} - 4 (x ^ {3} e ^ {x} - 3 x ^ {2} e ^ {x} + 6 x e ^ {x} - 6 e ^ {x} - 6 C)\) \(= x ^ {4} e ^ {x} - 4 x ^ {3} e ^ {x} + 1 2 x ^ {2} e ^ {x} - 2 4 x e ^ {x} + 2 4 e ^ {x} + 2 4 C\).

Notice how the constant of integration propagates. For a definite integral with a reduction formula, you’d apply the limits at each step or at the final evaluation. This example shows the power of breaking down a complex problem into simpler, recursive steps.

Other Reduction Formulae

Reduction formulas are not limited to \(\int x^n e^x dx\). They are a powerful tool for various forms of integrals.

More generally, we may let \(I_{n}\) be an indefinite integral or definite integral of the form:

• \(\int x (\ln x)^n dx\)
• \(\int \cos^n x dx\)
• \(\int \tan^n x dx\)
• \(\int \sec^n x dx\)
• \(\int_ {0} ^ {8} (x + 1) ^ {n} e ^ {2 x} d x\)

If we can express \(I_{n}\) in terms of \(I_{m}\), where \(m < n\), the expression obtained is known as a reduction formula for \(I_{n}\).

The general idea is to use integration by parts (or sometimes other methods) to express an integral involving a power \(n\) in terms of the same integral with a lower power, say \(n-1\) or \(n-2\). This allows for systematic evaluation.

6.5 Integration of Rational Functions

We now consider integrals of rational functions, which are ratios of polynomials.

We shall consider integrals like: \[ \int \frac {x ^ {3} + 3 x ^ {2}}{x ^ {2} + 1} d x, \quad \int \frac {2 + 3 x + x ^ {2}}{x (x ^ {2} + 1)} d x, \dots , \] or, in general, integrals of the form: \[ \int \frac {P (x)}{Q (x)} d x \] where \(P(x)\) and \(Q(x)\) are polynomials.

How do we handle integrals like \(\int {\frac {x + 2}{x ^ {3} - x}} d x\)?

This is done by writing \(\frac{P(x)}{Q(x)}\) as a sum of simpler fractions called “partial fractions”.

Integrating rational functions directly is often very difficult. Partial fraction decomposition breaks down the complex rational function into a sum of simpler fractions that are much easier to integrate. This is a very algebraic technique.

General Rational Functions

Before we can use partial fractions, we need to understand the factorization of polynomials.

Consider a rational function \(\int \frac{P(x)}{Q(x)} dx\) where \(P(x)\) and \(Q(x)\) are polynomials.

Pre-requisite: If \(\deg P(x) \geq \deg Q(x)\), first perform polynomial long division to write \(\frac{P(x)}{Q(x)}\) as a polynomial plus a proper rational function (where the degree of the numerator is less than the degree of the denominator).

Fact from Algebra: Every polynomial \(Q(x) = a _ {n} x ^ {n} + a _ {n - 1} x ^ {n - 1} + \dots + a _ {1} x + a _ {0}\) with real coefficients can be factorized as a product of:

• Linear factors: \((ax + b)\)
• Irreducible quadratic factors: \((ax^2 + bx + c)\) where \(b^2 - 4ac < 0\) (meaning it has no real roots).

The first step for integrating rational functions is always to check the degrees of the numerator and denominator. If the numerator’s degree is higher or equal, perform long division first. Then, you factorize the denominator completely into linear and irreducible quadratic factors. This factorization is critical for setting up the partial fraction decomposition.

Partial Fractions: General Forms

Based on the factors of \(Q(x)\), the corresponding partial fractions take specific forms:

Factors of \(Q(x)\)	Corresponding Partial Fractions
\((ax+b)^k\)	\(\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_k}{(ax+b)^k}\)
\((ax^2+bx+c)^k\)	\(\frac{A_1x+B_1}{ax^2+bx+c} + \dots + \frac{A_kx+B_k}{(ax^2+bx+c)^k}\)

Case I: Simple Linear Factors

If the denominator \(Q(x)\) is a product of distinct linear (degree 1) factors (say \(ax + b\)), then the corresponding partial fraction representation for each factor \(ax + b\) is \(\frac{A}{ax + b}\).

Example 6.5.1: Express \(\frac{1}{x(x - 1)}\) as partial fractions.

Solution:

There are two distinct linear factors \(x\) and \((x - 1)\) in the denominator.

Set up the partial fraction decomposition: \[ \frac {1}{x (x - 1)} = \frac {A}{x} + \frac {B}{x - 1}. \] Combine the right side: \[ {\frac {1}{x (x - 1)}} = {\frac {A (x - 1) + B x}{x (x - 1)}} = {\frac {(A + B) x - A}{x (x - 1)}}. \] Compare numerators: \(1 = (A + B) x - A\).

This equation must hold for all \(x\).

Comparing coefficients of \(x\): \(A + B = 0\).

Comparing constant terms: \(-A = 1 \quad \Rightarrow \quad A = -1\).

Substitute \(A = -1\) into \(A + B = 0 \quad \Rightarrow \quad -1 + B = 0 \quad \Rightarrow \quad B = 1\).

Hence, \(\frac{1}{x(x - 1)} = \frac{-1}{x} + \frac{1}{x - 1} = \frac{1}{x - 1} - \frac{1}{x}\).

This table summarizes the types of partial fractions. The first case, distinct linear factors, is the simplest. The goal is to set up the decomposition, combine the terms on the right, and then equate coefficients of like powers of x in the numerator to solve for the unknown constants A, B, etc.

Case I: Example Integration

Now let’s integrate a rational function with distinct linear factors.

Example 6.5.2: Evaluate \(\int \frac{x + 2}{x(x - 1)(x + 1)} dx\).

Solution:

1. Partial Fraction Decomposition:

   We first express \(\frac{P(x)}{Q(x)} = \frac{x + 2}{x(x - 1)(x + 1)}\) in partial fractions.

   There are three distinct linear factors in \(Q(x)\): \(x\), \((x - 1)\) and \((x + 1)\).

   Set up the decomposition: \[ \frac {x + 2}{x (x - 1) (x + 1)} = \frac {A}{x} + \frac {B}{x - 1} + \frac {C}{x + 1}. \] Combine the right side: \[ = \frac {A (x + 1) (x - 1) + B x (x + 1) + C x (x - 1)}{x (x - 1) (x + 1)}. \] Compare numerators: \(x + 2 = A (x^2 - 1) + B (x^2 + x) + C (x^2 - x)\). \(x + 2 = (A+B+C)x^2 + (B-C)x - A\).

Case I: Example Integration (cont.)

2. Solve for constants A, B, C:

   Comparing coefficients on the numerators:

   • Coefficient of \(x^2\): \(0 = A + B + C\)
   • Coefficient of \(x\): \(1 = B - C\)
   • Constant term (\(x^0\)): \(2 = - A \quad \Rightarrow \quad A = -2\).

   Substitute \(A=-2\) into the first equation: \(-2 + B + C = 0 \quad \Rightarrow \quad B + C = 2\).

   Now we have a system for \(B\) and \(C\):

   \(B - C = 1\)

   \(B + C = 2\)

   Adding the two equations: \(2B = 3 \quad \Rightarrow \quad B = 3/2\).

   Substitute \(B=3/2\) into \(B - C = 1 \quad \Rightarrow \quad 3/2 - C = 1 \quad \Rightarrow \quad C = 1/2\).

   So, \(A = \underline{-2}, \quad B = \underline{3/2}, \quad C = \underline{1/2}\).

Case I: Example Integration (cont.)

3. Integrate the partial fractions: \[ {\frac {x + 2}{x (x - 1) (x + 1)}} = {\frac {- 2}{x}} + {\frac {3 / 2}{x - 1}} + {\frac {1 / 2}{x + 1}}. \] \[ \int \frac {x + 2}{x (x - 1) (x + 1)} d x = \int \left(\frac {- 2}{x} + \frac {3 / 2}{x - 1} + \frac {1 / 2}{x + 1}\right) d x \] \[ = - 2 \int \frac {1}{x} d x + \frac {3}{2} \int \frac {1}{x - 1} d x + \frac {1}{2} \int \frac {1}{x + 1} d x \] \[ = - 2 \ln | x | + \frac {3}{2} \ln | x - 1 | + \frac {1}{2} \ln | x + 1 | + C. \]

This example shows the full process: decomposition into partial fractions, solving for coefficients, and then integrating each simple fraction (which typically results in logarithms). An alternative to equating coefficients is the “Heaviside cover-up method” for distinct linear factors, which can be faster.

Case II: Repeated Linear Factors

If \(Q(x)\) has repeated linear factors, the partial fraction decomposition is slightly more involved.

If \(Q(x)\) is a product of linear factors, some of which are repeated, say \((ax + b)^k\), where \(k \geq 2\), then there are \(k\) corresponding partial fractions: \[ \frac {A _ {1}}{a x + b}, \frac {A _ {2}}{(a x + b) ^ {2}}, \dots , \frac {A _ {k - 1}}{(a x + b) ^ {k - 1}}, \frac {A _ {k}}{(a x + b) ^ {k}} \]

Case II: Repeated Linear Factors

Example 6.5.3: Evaluate \(\int \frac{x^2}{(x - 3)(x + 2)^2} dx\).

Solution:

1. Partial Fraction Decomposition:

   For \((x - 3)\), the partial fraction is \(\frac{A}{x - 3}\).

   For \((x + 2)^2\), the partial fractions are \(\frac{B}{x + 2} + \frac{C}{(x + 2)^2}\).

   Set up the decomposition: \[ \frac {x ^ {2}}{(x - 3) (x + 2) ^ {2}} = \frac {A}{x - 3} + \frac {B}{x + 2} + \frac {C}{(x + 2) ^ {2}}. \] Combine the right side: \[ = \frac {A (x + 2) ^ {2} + B (x - 3) (x + 2) + C (x - 3)}{(x - 3) (x + 2) ^ {2}}. \] Compare numerators: \(x^2 = A(x^2 + 4x + 4) + B(x^2 - x - 6) + C(x - 3)\). \(x^2 = (A+B)x^2 + (4A-B+C)x + (4A-6B-3C)\).

Case II: Repeated Linear Factors (cont.)

2. Solve for constants A, B, C:

   Comparing coefficients:

   • \(x^2\): \(1 = A + B\)
   • \(x\): \(0 = 4A - B + C\)
   • Constant: \(0 = 4A - 6B - 3C\)

   From \(1 = A+B\), we get \(B = 1-A\). Substitute this into the other two equations:

   \(0 = 4A - (1-A) + C \quad \Rightarrow \quad 5A + C = 1 \quad \text{ (Eq. 1)}\)

   \(0 = 4A - 6(1-A) - 3C \quad \Rightarrow \quad 4A - 6 + 6A - 3C = 0 \quad \Rightarrow \quad 10A - 3C = 6 \quad \text{ (Eq. 2)}\)

   Multiply Eq. 1 by 3: \(15A + 3C = 3\).

   Add this to Eq. 2: \((15A + 3C) + (10A - 3C) = 3 + 6 \quad \Rightarrow \quad 25A = 9 \quad \Rightarrow \quad A = 9/25\).

   From Eq. 1: \(C = 1 - 5A = 1 - 5(9/25) = 1 - 9/5 = -4/5\).

   From \(B = 1-A\): \(B = 1 - 9/25 = 16/25\).

   So, \(A = \underline{9/25}, \quad B = \underline{16/25}, \quad C = \underline{-4/5}\).

Case II: Repeated Linear Factors (cont.)

3. Integrate the partial fractions: \[ \frac {x ^ {2}}{(x - 3) (x + 2) ^ {2}} = \frac {9/25}{x - 3} + \frac {16/25}{x + 2} - \frac {4/5}{(x + 2) ^ {2}}. \] \[ \int \frac {x ^ {2}}{(x - 3) (x + 2) ^ {2}} d x = \int \left(\frac {9}{25(x - 3)} + \frac {16}{25(x + 2)} - \frac {4}{5(x + 2) ^ {2}}\right) d x \] \[ = \frac {9}{25} \ln | x - 3 | + \frac {16}{25} \ln | x + 2 | - \frac {4}{5} \int (x + 2)^{-2} d x \] \[ = \frac {9}{25} \ln | x - 3 | + \frac {16}{25} \ln | x + 2 | - \frac {4}{5} \left( \frac{(x+2)^{-1}}{-1} \right) + C \] \[ = \frac {9}{25} \ln | x - 3 | + \frac {16}{25} \ln | x + 2 | + \frac {4}{5(x+2)} + C. \]

For repeated linear factors, you need a term for each power of the factor, up to the highest power. The algebra to solve for the constants can be more involved, often requiring solving a system of linear equations. The integration of \((x+2)^{-2}\) is a simple power rule.

Irreducible Quadratic Factors

Now we consider quadratic factors that cannot be factored into real linear terms.

The quadratic expression \(ax^2 + bx + c\) is said to be irreducible when its discriminant \(b^{2} - 4ac < 0\).

This means it has no real roots and cannot be factored further over real numbers.

We can often express \(ax^2 + bx + c\) in the form \(A(x+B)^2 + C^2\) by completing the square. This is useful for integration.

Case III: Distinct Irreducible Quadratic Factors

Suppose \(Q(x)\) contains the quadratic factor \(ax^2 + bx + c\) (\(b^2 - 4ac < 0\)). Then the partial fraction representation of \(\frac{P(x)}{Q(x)}\) will contain the term: \[ \frac {A x + B}{a x ^ {2} + b x + c} \] Note the numerator is linear \((Ax+B)\).

When you have an irreducible quadratic factor in the denominator, its corresponding partial fraction has a linear numerator (\(Ax+B\)). This is different from linear factors, which just have a constant numerator. Completing the square is a crucial algebraic step for integrating these types of terms.

A Useful Formula for Irreducible Quadratics

Integrating terms with irreducible quadratic denominators often involves the inverse tangent function.

Important

Theorem 6.5.4: \[ \int \frac{1}{a^2 + x^2} dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C. \]

A Useful Formula for Irreducible Quadratics

Example 6.5.5: \(\int \frac{1}{9x^2 + 25} dx\).

Solution:

We want to get this into the form \(\int \frac{1}{x^2 + k^2} dx\).

Factor out 9 from the denominator: \[ \int \frac {1}{9 x ^ {2} + 2 5} d x = \frac {1}{9} \int \frac {1}{x ^ {2} + \frac {25}{9}} d x \] Now we have the form \(\frac{1}{x^2 + a^2}\) where \(a^2 = \frac{25}{9}\), so \(a = \frac{5}{3}\). Applying the formula: \[ = \frac {1}{9} \left(\frac {1}{5 / 3} \tan^ {- 1} \frac {x}{5 / 3}\right) + C \] \[ = \frac {1}{9} \left(\frac {3}{5} \tan^ {- 1} \frac {3 x}{5}\right) + C = \frac {1}{1 5} \tan^ {- 1} \frac {3 x}{5} + C. \]

This specific formula for \(\int \frac{1}{a^2+x^2} dx\) is a direct result of the derivative of \(\tan^{-1}(x/a)\). You’ll encounter it frequently, so it’s good to memorize. The key is to algebraically manipulate the denominator to match the form \(x^2 + a^2\).

Completing the Square for Integration

Sometimes, we need to complete the square before applying the inverse tangent formula.

Example 6.5.6: Evaluate \(\int \frac{1}{x^2 + 4x + 5} dx\).

Solution:

1. Check for irreducibility: The discriminant \(b^{2} - 4ac = 4^2 - 4(1)(5) = 16 - 20 = -4 < 0\). So it’s irreducible.

2. Complete the square: \(x^2 + 4x + 5 = (x^2 + 4x + 4) + 1 = (x + 2)^2 + 1\).

3. Substitute and integrate: \[ \int \frac {1}{x ^ {2} + 4 x + 5} d x = \int \frac {1}{(x + 2) ^ {2} + 1} d x. \] Let \(u = x + 2\). Then \(du = dx\). \[ \int {\frac {1}{u ^ {2} + 1}} d u \]

   This matches the form \(\int \frac{1}{u^2 + a^2} du\) with \(a=1\).

   \(= \tan^ {- 1} u + C\)

   Substitute back \(u = x + 2\):

   \(= \tan^ {- 1} (x + 2) + C\).

Completing the Square for Integration

Example 6.5.7: Evaluate \(\int \frac{1}{4x^2 + 4x + 26} dx\).

Solution:

1. Complete the square:

   \(4x^2 + 4x + 26 = 4(x^2 + x) + 26 = 4(x^2 + x + 1/4 - 1/4) + 26\)

   \(= 4\left((x + 1/2)^2 - 1/4\right) + 26 = 4(x + 1/2)^2 - 1 + 26\)

   \(= 4(x + 1/2)^2 + 25 = (2(x + 1/2))^2 + 25 = (2x + 1)^2 + 5^2\).

2. Substitute and integrate: \[ \int {\frac {1}{4 x ^ {2} + 4 x + 2 6}} d x = \int {\frac {1}{(2 x + 1) ^ {2} + 5 ^ {2}}} d x. \] Let \(u = 2x + 1\). Then \(du = 2 dx\), so \(dx = \frac{1}{2} du\). \[ = \int {\frac {1}{u ^ {2} + 5 ^ {2}}} \left(\frac {1}{2} du\right) = \frac {1}{2} \int {\frac {1}{u ^ {2} + 5 ^ {2}}} d u \] Using \(\int \frac{1}{a^2 + u^2} du = \frac{1}{a}\tan^{-1}\frac{u}{a} + C\) with \(a=5\): \[ = \frac {1}{2} \left(\frac {1}{5} \tan^ {- 1} \frac {u}{5}\right) + C = \frac {1}{10} \tan^ {- 1} \frac {2x + 1}{5} + C. \]

These examples show that completing the square is a critical first step for integrating functions with irreducible quadratic denominators. You’re aiming to transform it into the \(\int \frac{1}{u^2+a^2} du\) form, possibly with a u-substitution.

Another Useful Integral Form

Sometimes, the numerator of a rational function with an irreducible quadratic denominator is a linear term that is related to the derivative of the denominator.

Important

Theorem 6.5.8: \[ \int \frac{2ax + b}{ax^2 + bx + c} dx = \ln |ax^2 +bx + c| + C \] This comes from the general rule: \(\int \frac {f ^ {\prime} (x)}{f (x)} d x = \ln | f (x) | + C\).

Another Useful Integral Form

Example 6.5.9: \(\int \frac{2x + 1}{4x^2 + 4x + 26} dx\).

Solution:

The derivative of the denominator \(f(x) = 4x^2 + 4x + 26\) is \(f'(x) = 8x + 4\).

The numerator is \(2x+1\). We can adjust it to match \(8x+4\):

\(2x+1 = \frac{1}{4}(8x+4)\).

So, \[ \int {\frac {2 x + 1}{4 x ^ {2} + 4 x + 2 6}} d x = \int {\frac {\frac{1}{4}(8 x + 4)}{4 x ^ {2} + 4 x + 2 6}} d x \] \[ = \frac {1}{4} \int \frac {8 x + 4}{4 x ^ {2} + 4 x + 2 6} d x \] Now the numerator is exactly the derivative of the denominator. \[ = \frac {1}{4} \ln \left| 4 x ^ {2} + 4 x + 2 6 \right| + C. \]

(Since \(4x^2+4x+26\) is always positive, we can drop the absolute value).

This is a specific case of u-substitution where \(u\) is the entire denominator. If the numerator is a constant multiple of the derivative of the denominator, the integral is a logarithm. This is a very common pattern to look for.

Combining Techniques

Often, you’ll need to combine multiple techniques for one integral.

Example 6.5.10: Evaluate \(\int {\frac {x}{x ^ {2} + 4 x + 1 3}} d x\).

Solution:

1. Check irreducibility and complete the square:

   Discriminant \(b^{2} - 4ac = 4^{2} - 4(1)(13) = 16 - 52 = -36 < 0\), so it’s irreducible.

   Completing the square: \(x^{2} + 4x + 13 = (x + 2)^{2} + 9\).

2. Split the numerator: The numerator is \(x\). The derivative of the denominator is \(2x+4\). We want to express the numerator in terms of \(2x+4\).

   \(x = A(2x+4) + B\).

   Comparing coefficients of \(x\): \(1 = 2A \quad \Rightarrow \quad A = 1/2\).

   Comparing constant terms: \(0 = 4A + B \quad \Rightarrow \quad 0 = 4(1/2) + B \quad \Rightarrow \quad 0 = 2 + B \quad \Rightarrow \quad B = -2\).

   So, \(x = \frac{1}{2}(2x+4) - 2\).

   Rewrite the integral: \[ \int {\frac {x}{x ^ {2} + 4 x + 1 3}} d x = \int {\frac {\frac{1}{2}(2 x + 4) - 2}{x ^ {2} + 4 x + 1 3}} d x \] \[ = \frac {1}{2} \int {\frac {2 x + 4}{x ^ {2} + 4 x + 1 3}} d x - 2 \int {\frac {1}{x ^ {2} + 4 x + 1 3}} d x. \]

Combining Techniques (cont.)

3. Integrate each part:

   • First integral (logarithmic form):

     \(\frac {1}{2} \int {\frac {(2 x + 4)}{x ^ {2} + 4 x + 1 3}} d x = \frac{1}{2} \ln |x^2 + 4x + 13| + C_1\).

   • Second integral (inverse tangent form, using completed square):

     \(2 \int {\frac {1}{x ^ {2} + 4 x + 1 3}} d x = 2 \int {\frac {1}{(x + 2) ^ {2} + 3 ^ {2}}} d x\).

     Let \(u = x+2\), \(du=dx\). This becomes \(2 \int \frac{1}{u^2+3^2}du = 2 \cdot \frac{1}{3}\tan^{-1}\left(\frac{u}{3}\right) + C_2 = \frac{2}{3}\tan^{-1}\left(\frac{x+2}{3}\right) + C_2\).

4. Combine the results: \[ \int {\frac {x}{x ^ {2} + 4 x + 1 3}} d x = \frac {1}{2} \ln | x ^ {2} + 4 x + 1 3 | - \frac{2}{3}\tan^{-1}\left(\frac{x+2}{3}\right) + C. \]

This is a standard approach for linear over irreducible quadratic. You split the numerator so that one part is the derivative of the denominator (for a logarithm) and the other is a constant (for an inverse tangent).

Combining Factors: Linear and Irreducible Quadratic

When the denominator has both linear and irreducible quadratic factors, you combine the partial fraction types.

Example 6.5.11: Evaluate \(\int \frac{x - 1}{x^3 + x} dx\).

Solution:

1. Factor the denominator: \(x^{3} + x = x(x^{2} + 1)\). Here, \(x\) is a linear factor and \(x^{2} + 1\) is an irreducible quadratic factor (\(b^2-4ac = 0^2-4(1)(1) = -4 < 0\)).
2. Partial Fraction Decomposition: \[ {\frac {x - 1}{x ^ {3} + x}} = {\frac {A}{x}} + {\frac {B x + C}{x ^ {2} + 1}}. \] Combine the right side: \[ = {\frac {A (x ^ {2} + 1) + (B x + C) x}{x (x ^ {2} + 1)}} = {\frac {A x ^ {2} + A + B x ^ {2} + C x}{x (x ^ {2} + 1)}} = {\frac {(A + B) x ^ {2} + C x + A}{x (x ^ {2} + 1)}}. \] Compare numerators: \(x - 1 = (A + B) x ^ {2} + C x + A\).

Combining Factors: Linear and Irreducible Quadratic (cont.)

3. Solve for constants A, B, C:
   • \(x^2\): \(0 = A + B\)
   • \(x\): \(1 = C\)
   • Constant: \(-1 = A\) From \(A = -1\) and \(A+B=0 \quad \Rightarrow \quad -1+B=0 \quad \Rightarrow \quad B = 1\). So, \(A = -1, B = 1, C = 1\).
4. Integrate the partial fractions: \[ \int {\frac {x - 1}{x ^ {3} + x}} d x = \int \left(-\frac {1}{x} + \frac {x + 1}{x ^ {2} + 1}\right) d x = - \int {\frac {1}{x}} d x + \int {\frac {x + 1}{x ^ {2} + 1}} d x. \]
   • First integral: \(- \int {\frac {1}{x}} d x = - \ln | x | + C_1\).
   • Second integral: Split \(\int {\frac {x + 1}{x ^ {2} + 1}} d x = \int \left(\frac{x}{x^2+1} + \frac{1}{x^2+1}\right) dx\). \(\int \frac{x}{x^2+1} dx\): Let \(u=x^2+1, du=2x dx\). So \(\frac{1}{2}\int \frac{1}{u}du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+1)\). \(\int \frac{1}{x^2+1} dx = \tan^{-1}x\). So, \(\int {\frac {x + 1}{x ^ {2} + 1}} d x = \frac {1}{2} \ln (x ^ {2} + 1) + \tan^ {- 1} x + C_2\).
5. Combine the results: \[ \int {\frac {x - 1}{x ^ {3} + x}} d x = - \ln | x | + \frac {1}{2} \ln (x ^ {2} + 1) + \tan^ {- 1} x + C. \]

This example perfectly demonstrates the need to combine the two types of partial fractions (linear factor and irreducible quadratic factor) and then apply the specific integration techniques for each. It’s common to split the numerator of the irreducible quadratic term.

Case IV: Repeating Irreducible Factors (Optional)

This case is more complex and often covered in advanced courses.

Suppose \(Q(x)\) contains the repeating irreducible quadratic factor \((ax^2 + bx + c)^k\). Then the partial fraction representation of \(\frac{P(x)}{Q(x)}\) will contain the terms: \[ \frac {A _ {1} x + B _ {1}}{a x ^ {2} + b x + c} + \frac {A _ {2} x + B _ {2}}{(a x ^ {2} + b x + c) ^ {2}} + \dots + \frac {A _ {k} x + B _ {k}}{(a x ^ {2} + b x + c) ^ {k}} \]

Example 6.5.12: Evaluate \(\int \frac{x^2}{x(x^2 + 4)^3} dx\).

Solution:

Partial fractions: \[ {\frac {x ^ {2}}{x (x ^ {2} + 4) ^ {3}}} = {\frac {A}{x}} + {\frac {A _ {1} x + B _ {1}}{x ^ {2} + 4}} + {\frac {A _ {2} x + B _ {2}}{(x ^ {2} + 4) ^ {2}}} + {\frac {A _ {3} x + B _ {3}}{(x ^ {2} + 4) ^ {3}}} \] Proceed like the above examples to solve for \(A_{i}, B_{i}\) and \(A\), and take care of each partial fraction. The integration of terms like \(\frac{Ax+B}{(x^2+C)^k}\) for \(k \ge 2\) can be very involved.

While the setup is logical (a term for each power), the algebra to solve for the coefficients and the integration itself become significantly more complicated for repeated irreducible quadratic factors. This is usually where computational tools or integral tables become essential.

Degree of Numerator \(\geq\) Degree of Denominator

Remember the first step: if \(\deg P(x) \geq \deg Q(x)\), use polynomial long division.

Example 6.5.13: Evaluate \(\int {\frac {x ^ {3} + 3 x ^ {2}}{x ^ {2} + 1}} d x\).

Solution:

1. Perform long division: \[ \begin{array}{r} x+3 \\ x^2+1 \overline{)x^3+3x^2+0x+0} \\ -(x^3+0x^2+x) \\ \hline 3x^2-x+0 \\ -(3x^2+0x+3) \\ \hline -x-3 \end{array} \] So, \(\frac {x ^ {3} + 3 x ^ {2}}{x ^ {2} + 1} = x + 3 + {\frac {- x - 3}{x ^ {2} + 1}}\).

Degree of Numerator \(\geq\) Degree of Denominator (cont.)

2. Rewrite and integrate: \[ \int \frac {x ^ {3} + 3 x ^ {2}}{x ^ {2} + 1} d x = \int \left(x + 3 - {\frac {x}{x ^ {2} + 1}} - {\frac {3}{x ^ {2} + 1}}\right) d x. \] Integrate term by term:
   • \(\int (x + 3) d x = \frac {x ^ {2}}{2} + 3 x + C _ {1}\).
   • \(\int \frac {x}{x ^ {2} + 1} d x\): Let \(u=x^2+1\), \(du=2x dx\). So \(\frac{1}{2}\int \frac{1}{u}du = \frac{1}{2}\ln(x^2+1) + C_2\).
   • \(\int \frac {3}{x ^ {2} + 1} d x = 3 \int \frac {1}{x ^ {2} + 1} d x = 3 \tan^{-1} x + C_3\).
3. Combine the results: \[ \int {\frac {x ^ {3} + 3 x ^ {2}}{x ^ {2} + 1}} d x = \frac {x ^ {2}}{2} + 3 x - \frac {1}{2} \ln (x ^ {2} + 1) - 3 \tan^ {- 1} x + C. \]

Another example:

\(\int \frac {x ^ {3} + 2}{x ^ {3} - x} d x = \int \left(1 + \frac {x + 2}{x ^ {3} - x}\right) d x = x + \int \frac {x + 2}{x (x - 1) (x + 1)} d x.\)

The last integral can be handled with the method of partial fractions (as in Example 6.5.2).

Long division transforms an improper rational function into a polynomial (easy to integrate) and a proper rational function (which you then tackle with partial fractions). This is always the first step for this type of problem.

6.6 Improper Integrals

So far, we’ve assumed our integrals have a continuous integrand over a closed and bounded interval. But what if they don’t?

We know that a continuous function is Riemann integrable over a closed and bounded interval \([a,b]\) (i.e., the integral \(\int_{a}^{b}f(x)dx\) has a finite value). More generally, if \(f\) is not continuous at a finite number of points in \([a,b]\), then \(\int_{a}^{b}f(x)dx\) exists (as long as these discontinuities are “jump” type and the function remains bounded).

However, when the integrand \(f\) is not bounded on \([a,b]\) or the interval \([a,b]\) is no longer bounded, we have to consider the corresponding integrals carefully. The integral may or may not converge to a finite value. Such integrals are known as improper integrals.

We shall discuss two types of improper integrals.

Improper integrals extend the concept of integration to cases where the function might go to infinity or the interval itself goes to infinity. The idea is to use limits to define these integrals.

Type 1: Unbounded Integrand

This type occurs when the function itself becomes infinite at some point within the interval of integration.

Consider the integral \(\int_ {- 1} ^ {1} \frac {1}{x ^ {2}} d x\).

This integral is not defined in the Riemann sense on \([-1, 1]\), because \(\frac{1}{x^2}\) is not bounded on \([-1, 1]\) since \(\lim_{x \to 0} \frac{1}{x^2} = \infty\). The point \(x=0\) is called a singular point.

For the same reason, neither \(\int_ {- 1} ^ {0} \frac {1}{x ^ {2}} d x\) nor \(\int_ {0} ^ {1} \frac {1}{x ^ {2}} d x\) are defined as standard Riemann integrals.

Type 2: Unbounded Interval

This type occurs when one or both limits of integration are infinite.

Consider the integral \(\int_{1}^{\infty}\frac{x}{1 + x^2} dx\) and \(\int_{-\infty}^{-1}\frac{1}{x^3} dx\). Both intervals \([1, \infty)\) or \((- \infty, -1]\) are not bounded.

It’s important to recognize these two distinct types of improper integrals. For both, the approach will involve replacing the problematic part with a variable and taking a limit.

6.6.1 Improper Integral (Type 1): Unbounded Integrand

Let’s evaluate an improper integral with an unbounded integrand.

Example 6.6.1: Evaluate \(\int_{-1}^{0}\frac{1}{x^2} dx\).

Solution:

Since \(\lim_{x\to 0^{-}}\frac{1}{x^2} = \infty\), the point \(x=0\) is a singular point.

To handle this, we replace the upper limit 0 by a variable \(t < 0\) and take a limit: \[ \int_ {- 1} ^ {0} \frac {1}{x ^ {2}} d x = \lim _ {t \to 0 ^ {-}} \int_ {- 1} ^ {t} \frac {1}{x ^ {2}} d x. \] Now evaluate the proper integral: \(\int_ {- 1} ^ {t} \frac {1}{x ^ {2}} d x = \int_ {- 1} ^ {t} x^{-2} d x = \left[ \frac{x^{-1}}{-1} \right]_{-1}^t = \left[ -\frac{1}{x} \right]_{-1}^t\) \(= \left(-\frac{1}{t}\right) - \left(-\frac{1}{-1}\right) = -\frac{1}{t} - 1\). Now take the limit: \[ = \lim _ {t \rightarrow 0 ^ {-}} \left(-\frac{1}{t} - 1\right). \] As \(t \to 0^{-}\) (approaching 0 from the left, meaning \(t\) is a small negative number), \(\frac{1}{t} \to -\infty\). So, \(-\frac{1}{t} \to \infty\).

Therefore, the limit is \(\infty - 1 = \infty\).

We say that the improper integral \(\int_{-1}^{0}\frac{1}{x^2} dx\) diverges to infinity.

The key step is replacing the singular point with a variable (t) and then taking the limit. If this limit exists and is finite, the integral converges; otherwise, it diverges. Here, the limit goes to infinity, so it diverges.

Improper Integral (Type 1): More Examples

Let’s look at another example of an unbounded integrand.

Example 6.6.2: Evaluate \(\int_0^1\frac{1}{\sqrt{2x - x^2}} dx\).

Solution:

1. Identify singular points: The denominator is \(\sqrt{x(2-x)}\). At \(x=0\), the denominator is 0, so \(\lim_{x\to 0^{+}}\frac{1}{\sqrt{2x - x^2}} = \infty\). \(x=0\) is a singular point. At \(x=1\), the denominator is \(\sqrt{1(1)} = 1\), no singularity.
2. Set up the limit: \[ \int_ {0} ^ {1} \frac {1}{\sqrt {2 x - x ^ {2}}} d x = \lim _ {t \rightarrow 0 ^ {+}} \left(\int_ {t} ^ {1} \frac {1}{\sqrt {2 x - x ^ {2}}} d x\right). \]
3. Find the antiderivative (complete the square in the denominator): \(2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = -((x - 1)^2 - 1) = 1 - (x - 1)^2\). So the integral is \(\int \frac{1}{\sqrt{1 - (x - 1)^2}} dx\). This is the form of \(\sin^{-1}(x-1)\). \(\int_ {t} ^ {1} \frac {1}{\sqrt {1 - (x - 1) ^ {2}}} d x = \left[ \sin^{-1}(x - 1) \right]_t^1\) \(= \sin^{-1}(1 - 1) - \sin^{-1}(t - 1) = \sin^{-1}(0) - \sin^{-1}(t - 1) = 0 - \sin^{-1}(t - 1) = -\sin^{-1}(t - 1)\).
4. Take the limit: \[ = \lim _ {t \rightarrow 0 ^ {+}} \left(-\sin^ {- 1} (t - 1)\right) \] As \(t \to 0^{+}\), \((t-1) \to -1\). So, \(= -\sin^{-1}(-1) = -(-\pi/2) = \pi/2\). We say that the improper integral \(\int_0^1\frac{1}{\sqrt{2x - x^2}} dx\) converges to \(\pi/2\).

Here, the integral converges to a finite value. This shows that improper integrals can indeed have finite values. Completing the square was key to finding the antiderivative as an inverse sine function.

Improper Integral (Type 1): Multiple Singular Points

If there are multiple singular points or a singular point within the interval, you must split the integral.

Example 6.6.3: Evaluate \(\int_0^2\frac{1}{\sqrt{2x - x^2}} dx\).

Solution:

1. Identify singular points: \(\lim_{x\to 0^{+}}\frac{1}{\sqrt{2x - x^2}} = \infty\) (singular point at \(x=0\)). \(\lim_{x\to 2^{-}}\frac{1}{\sqrt{2x - x^2}} = \infty\) (singular point at \(x=2\)). The function \(\frac{1}{\sqrt{2x - x^2}}\) is continuous on \((0,2)\).
2. Split the integral: We choose a point in the interval, e.g., \(x=1\). \[ \int_ {0} ^ {2} \frac {1}{\sqrt {2 x - x ^ {2}}} d x = \int_ {0} ^ {1} \frac {1}{\sqrt {2 x - x ^ {2}}} d x + \int_ {1} ^ {2} \frac {1}{\sqrt {2 x - x ^ {2}}} d x. \]
3. Evaluate each improper integral:
   • The integral \(\int_0^1\frac{1}{\sqrt{2x - x^2}} dx\) was evaluated in the preceding example and converges to \(\pi/2\).
   • Now evaluate \(\int_{1}^{2}\frac{1}{\sqrt{2x - x^2}} dx\): \[ \int_ {1} ^ {2} \frac {1}{\sqrt {2 x - x ^ {2}}} d x = \lim _ {t \rightarrow 2 ^ {-}} \left(\int_ {1} ^ {t} \frac {1}{\sqrt {1 - (x - 1) ^ {2}}} d x\right) \] \(= \lim _ {t \rightarrow 2 ^ {-}} \left[ \sin^ {- 1} (x - 1) \right]_1^t\) \(= \lim _ {t \rightarrow 2 ^ {-}} \left(\sin^ {- 1} (t - 1) - \sin^{-1}(1-1)\right)\) \(= \lim _ {t \rightarrow 2 ^ {-}} \left(\sin^ {- 1} (t - 1) - 0\right)\) As \(t \to 2^{-}\), \((t-1) \to 1\). \(= \sin^{-1}(1) = \pi/2\).
4. Add the results: \[ \int_ {0} ^ {2} \frac {1}{\sqrt {2 x - x ^ {2}}} d x = \frac{\pi}{2} + \frac{\pi}{2} = \pi. \] The improper integral converges to \(\pi\).

When an integrand has singularities at both ends of the interval, or even in the middle, you must split the integral at each singular point and evaluate each resulting improper integral separately. If any of the split integrals diverge, the original integral diverges.

6.6.2 Improper Integral (Type 2): Unbounded Intervals

Now we consider improper integrals over an unbounded interval.

\[ \int_ {a} ^ {\infty} f (x) d x, \int_ {- \infty} ^ {b} f (x) d x, \int_ {- \infty} ^ {\infty} f (x) d x \]

To integrate over an unbounded interval:

• Replace the infinite limit by a variable (e.g., \(t\)).
• Pass the definite integral to a limiting process: \(t \to \infty\) or \(t \to -\infty\).

For the improper integral of the form \(\int_{a}^{\infty}f(x)dx\), we define it as: \[ \int_ {a} ^ {\infty} f (x) d x = \lim _ {t \rightarrow \infty} \int_ {a} ^ {t} f (x) d x. \] If this limit exists and is finite, the integral converges; otherwise, it diverges.

The process is similar to Type 1: replace the problematic part (infinity) with a variable and take a limit. This transforms an improper integral into a proper definite integral that can be evaluated using FTC Part 2, followed by a limit calculation.

Improper Integral (Type 2): Examples

Let’s evaluate integrals with infinite limits.

Example 6.6.4: Evaluate \(\int_{1}^{\infty}\frac{1}{x^2} dx\).

Solution:

1. Set up the limit: \[ \int_{1}^{\infty}\frac{1}{x^{2}} dx = \lim_{t\to \infty}\int_{1}^{t}\frac{1}{x^{2}} dx. \]
2. Evaluate the proper integral: \(\int_{1}^{t}\frac{1}{x^{2}} dx = \left[ -\frac{1}{x} \right]_1^t = \left(-\frac{1}{t}\right) - \left(-\frac{1}{1}\right) = -\frac{1}{t} + 1\).
3. Take the limit: \[ = \lim _ {t \rightarrow \infty} \left(- \frac {1}{t} + 1\right). \] As \(t \to \infty\), \(\frac{1}{t} \to 0\). So, the limit is \(0 + 1 = 1\). Thus, the improper integral \(\int_{1}^{\infty}\frac{1}{x^2} dx\) converges to 1.

This is a convergent improper integral. The area under \(1/x^2\) from 1 to infinity is finite. This is a common result to remember.

Improper Integral (Type 2): More Examples

Let’s look at another example with an infinite lower limit.

For the improper integral of the form \(\int_{-\infty}^{b}f(x)dx\), we define it as: \[ \int_ {- \infty} ^ {b} f (x) d x = \lim _ {t \to - \infty} \int_ {t} ^ {b} f (x) d x. \]

Example 6.6.5: Evaluate \(\int_{-\infty}^{-1}\frac{1}{x^3} dx\).

Solution:

1. Set up the limit: \[ \int_{-\infty}^{-1}\frac{1}{x^3} dx = \lim_{t\to -\infty}\int_t^{-1}\frac{1}{x^3} dx. \]
2. Evaluate the proper integral: \(\int_t^{-1}\frac{1}{x^3} dx = \int_t^{-1} x^{-3} dx = \left[ \frac{x^{-2}}{-2} \right]_t^{-1} = \left[ -\frac{1}{2x^2} \right]_t^{-1}\). \(= \left(-\frac{1}{2(-1)^2}\right) - \left(-\frac{1}{2t^2}\right) = -\frac{1}{2} + \frac{1}{2t^2}\).
3. Take the limit: \[ = \lim _ {t \to - \infty} \left(- \frac {1}{2} + \frac {1}{2 t ^ {2}}\right). \] As \(t \to -\infty\), \(t^2 \to \infty\), so \(\frac{1}{2t^2} \to 0\). Therefore, the limit is \(-\frac{1}{2} + 0 = -\frac{1}{2}\). Thus, the improper integral \(\int_{-\infty}^{-1}\frac{1}{x^3} dx\) converges to \(-1/2\).

This example shows convergence for a negative result, which is perfectly fine for integrals representing net area. Remember the rules for evaluating limits as x approaches positive or negative infinity.

6.6.3 Comparison Test for Integrals

Often, we want to estimate whether a given improper integral is convergent or divergent without actually evaluating it. The comparison test is useful here.

Important

Theorem 6.6.6 (Comparison Theorem for Integrals):

Suppose \(f\) and \(g\) are continuous functions such that \(f(x) \geq g(x) \geq 0\), for \(x \geq a\). Then:

1. If \(\int_{a}^{\infty} f(x) \, dx\) converges, then \(\int_{a}^{\infty} g(x) \, dx\) converges. (If the larger integral is finite, the smaller one must also be finite.)

2. If \(\int_{a}^{\infty}g(x)dx\) diverges, then \(\int_{a}^{\infty}f(x)dx\) diverges.

   (If the smaller integral is infinite, the larger one must also be infinite.)

Important: Both functions \(f(x)\) and \(g(x)\) must be non-negative for this theorem to apply.

This theorem is analogous to the comparison test for series. It’s incredibly useful when an integral is too hard to evaluate directly. The key is to find a function that you can integrate and that has the right relationship (\(f \ge g \ge 0\)) to your original function.

Comparison Test: Example

Let’s use the comparison test to determine convergence or divergence.

Example 6.6.7: Determine whether \(\int_{2}^{\infty}\frac{1}{\ln x} dx\) converges or diverges.

Solution:

1. Find a comparison function: For \(x > 2\), we know that \(\ln x < x\). (To see this, consider \(h(x) = x - \ln x\). \(h(1)=1\), \(h'(x) = 1 - 1/x\). For \(x > 1\), \(h'(x)>0\), so \(h(x)\) is increasing. Thus \(x > \ln x\) for \(x > 1\).) Since \(\ln x < x\), then taking reciprocals (and assuming both are positive, which they are for \(x>2\)): \[ \frac {1}{\ln x} > \frac {1}{x}. \]

2. Evaluate the comparison integral: Let \(g(x) = \frac{1}{x}\) and \(f(x) = \frac{1}{\ln x}\). We have \(f(x) \geq g(x) \geq 0\) for \(x \geq 2\). Now, consider \(\int_{2}^{\infty} g(x) dx = \int_{2}^{\infty} \frac{1}{x} dx\). \[ \int_ {2} ^ {\infty} \frac {1}{x} d x = \lim_{t\to\infty} \int_2^t \frac{1}{x} dx = \lim_{t\to\infty} [\ln|x|]_2^t = \lim_{t\to\infty} (\ln t - \ln 2) = \infty. \] The integral \(\int_{2}^{\infty} \frac{1}{x} dx\) diverges.

3. Apply the Comparison Theorem: Since the smaller integral \(\int_{2}^{\infty} \frac{1}{x} dx\) diverges, by the Comparison Theorem (property 2), we conclude that \(\int_{2}^{\infty}\frac{1}{\ln x} dx\) also diverges.

This example perfectly demonstrates how to use the comparison test. We found a simpler function (\(1/x\)) that was smaller than our target function but also diverged. Since the “smaller” one diverged, the “larger” one must also diverge.

p-Integrals

The following improper integrals known as p-integrals are very useful in comparison tests because their convergence/divergence properties are well-known.

Important

Theorem 6.6.8 (p-integrals): Suppose \(a \in \mathbb{R}\) and \(a > 0\). Then:

1. For Type 2 improper integrals (unbounded interval):

   \[ \int_ {a} ^ {\infty} \frac {1}{x ^ {p}} d x = \left\{ \begin{array}{l l} \frac {a ^ {1 - p}}{p - 1} & \mathrm{i f} p > 1, \\ \infty & \mathrm{i f} p \leq 1. \end{array} \right. \]

2. For Type 1 improper integrals (unbounded integrand at 0):

   \[ \int_ {0} ^ {a} \frac {1}{x ^ {p}} d x = \left\{ \begin{array}{l l} \frac {a ^ {1 - p}}{1 - p} & \mathrm{i f} p < 1, \\ \infty & \mathrm{i f} p \geq 1. \end{array} \right. \]

These p-integral formulas are very handy for quick comparisons. Notice the conditions for convergence are opposite for Type 1 and Type 2 integrals. It’s often helpful to think of the area “closing in” or “spreading out” as x goes to infinity or approaches zero.

Proof of p-integral (a)

Let’s prove the first part of the p-integral theorem.

Proof of (a):

We need to evaluate \(\int_{a}^{\infty} \frac{1}{x^p} dx\).

Case 1: \(p = 1\).

\(\int_{a}^{\infty} \frac{1}{x} dx = \lim_{t\to\infty} \int_a^t \frac{1}{x} dx = \lim_{t\to\infty} [\ln|x|]_a^t = \lim_{t\to\infty} (\ln t - \ln a) = \infty\). So, for \(p=1\), the integral diverges.

Case 2: \(p \neq 1\).

\(\int_{a}^{\infty} \frac{1}{x^p} dx = \lim_{t\to\infty} \int_a^t x^{-p} dx\).

The antiderivative is \(\frac{x^{-p+1}}{-p+1}\). So, \(\lim_{t\to\infty} \left[ \frac{x^{1-p}}{1-p} \right]_a^t = \lim_{t\to\infty} \left( \frac{t^{1-p}}{1-p} - \frac{a^{1-p}}{1-p} \right)\).

Now, consider the behavior of \(t^{1-p}\) as \(t \to \infty\):

• If \(1-p < 0\) (i.e., \(p > 1\)): Then \(t^{1-p} = \frac{1}{t^{p-1}} \to 0\) as \(t \to \infty\). The limit becomes \(0 - \frac{a^{1-p}}{1-p} = \frac{a^{1-p}}{-(1-p)} = \frac{a^{1-p}}{p-1}\). (Converges)
• If \(1-p > 0\) (i.e., \(p < 1\)): Then \(t^{1-p} \to \infty\) as \(t \to \infty\). The limit becomes \(\infty - \frac{a^{1-p}}{1-p} = \infty\). (Diverges)

Combining both cases: \[ \int_ {a} ^ {\infty} \frac {1}{x ^ {p}} d x = \left\{ \begin{array}{l l} \frac {a ^ {1 - p}}{p - 1}, & \mathrm {i f} p > 1; \\ \infty , & \mathrm {i f} p \leq 1. \end{array} \right. \]

The proof relies on evaluating the integral directly using the power rule for integration and then analyzing the limit as t approaches infinity, considering different cases for p. This is a fundamental result for improper integrals.