Partial Fractions Method for Rational Functions | Integration Techniques

What You'll Learn

• General methods and theoretical foundations for rational function integration
• Four standard steps of partial fraction decomposition
• Three typical integral types and their solution methods
• 5 detailed examples from basic to advanced
• Key techniques and common pitfalls

1. Method Principles

Definition 1.1: Rational Functions

A rational function is a function of the form $R(x) = \frac{P(x)}{Q(x)}$ , where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$ .

Based on the relationship between numerator and denominator degrees, rational functions are classified as:

Proper fraction: $\deg(P) < \deg(Q)$
Improper fraction: $\deg(P) \geq \deg(Q)$

Core Idea

The essence of partial fractions is to decompose a complex rational function into a sum of simpler fractions, each of which can be integrated directly. Like a "divide and conquer" strategy, we break the whole into parts and tackle each individually.

Theorem 1.2: Partial Fraction Decomposition Theorem

Any proper fraction can be decomposed into a sum of the following three basic forms:

Type 1: Linear factor $(x-a)$

\frac{A}{x-a}

Type 2: Repeated linear factor $(x-a)^n$

\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}

Type 3: Irreducible quadratic factor $(x^2 + px + q)^n$ (where $\Delta = p^2 - 4q < 0$ )

\frac{B_1x + C_1}{x^2 + px + q} + \frac{B_2x + C_2}{(x^2 + px + q)^2} + \cdots + \frac{B_nx + C_n}{(x^2 + px + q)^n}

2. Standard Solution Steps

1

Determine Fraction Type

Compare the degrees of numerator and denominator. If it's an improper fraction ( $\deg(P) \geq \deg(Q)$ ), first use polynomial division to convert it to "polynomial + proper fraction" form.

2

Factor the Denominator

Factor the denominator $Q(x)$ into a product of linear and irreducible quadratic factors. This is a critical step that determines the form of the decomposition.

3

Set Up Undetermined Coefficients

Based on the factored form of the denominator, write the partial fraction decomposition and introduce undetermined coefficients $A_i, B_i, C_i$ , etc.

4

Solve for Coefficients and Integrate

After combining fractions, compare coefficients on both sides or substitute special values to find all undetermined coefficients. Then integrate each partial fraction separately.

3. Detailed Examples

Example 3.1: Basic: Simple Linear Factors

Problem: Find $\int \frac{x+3}{x^2+2x+4}\,dx$

Solution:

Step 1: Complete the Square

First, complete the square for the denominator:

x^2 + 2x + 4 = (x+1)^2 + 3 = (x+1)^2 + (\sqrt{3})^2

Step 2: Split the Numerator

Rewrite the numerator in a form related to the derivative of the denominator:

x + 3 = \frac{1}{2}(2x+2) + 2 = \frac{1}{2} \cdot \frac{d}{dx}(x^2+2x+4) + 2

Step 3: Decompose the Integral

\begin{aligned} I &= \int \frac{x+3}{x^2+2x+4}\,dx \\ &= \int \frac{\frac{1}{2}(2x+2) + 2}{x^2+2x+4}\,dx \\ &= \frac{1}{2}\int \frac{2x+2}{x^2+2x+4}\,dx + 2\int \frac{1}{(x+1)^2+(\sqrt{3})^2}\,dx \end{aligned}

Step 4: Calculate the Integral

The first term is a logarithmic integral, the second is an arctangent integral:

\begin{aligned} I &= \frac{1}{2}\ln|x^2+2x+4| + 2 \cdot \frac{1}{\sqrt{3}}\arctan\frac{x+1}{\sqrt{3}} + C \\ &= \frac{1}{2}\ln(x^2+2x+4) + \frac{2\sqrt{3}}{3}\arctan\frac{x+1}{\sqrt{3}} + C \end{aligned}

Answer

I = \frac{1}{2}\ln(x^2+2x+4) + \frac{2\sqrt{3}}{3}\arctan\frac{x+1}{\sqrt{3}} + C

Example 3.2: Partial Fractions with Repeated Factors

Problem: Find $\int \frac{1}{(a^2-x^2)^2}\,dx$

Solution:

Step 1: Factor the Denominator

Notice this is the square of a difference of squares:

(a^2-x^2)^2 = [(a-x)(a+x)]^2 = (a-x)^2(a+x)^2

Step 2: Partial Fraction Decomposition

Let:

\frac{1}{(a^2-x^2)^2} = \frac{A}{a-x} + \frac{B}{(a-x)^2} + \frac{C}{a+x} + \frac{D}{(a+x)^2}

Step 3: Find Undetermined Coefficients

After combining fractions, compare coefficients (using symmetry simplifies this):

By symmetry, $A = -C$ , $B = D$

Substituting $x = a$ : $\frac{1}{(2a)^2} = \frac{B}{0}$ , we get $B = \frac{1}{4a^2}$

Substituting $x = 0$ and solving, we get $A = \frac{1}{4a^3}$

Step 4: Calculate the Integral

\begin{aligned} I &= \int \left(\frac{1}{4a^3(a-x)} + \frac{1}{4a^2(a-x)^2} - \frac{1}{4a^3(a+x)} + \frac{1}{4a^2(a+x)^2}\right)dx \\ &= -\frac{1}{4a^3}\ln|a-x| + \frac{1}{4a^2(a-x)} + \frac{1}{4a^3}\ln|a+x| + \frac{1}{4a^2(a+x)} + C \\ &= \frac{1}{4a^3}\ln\left|\frac{a+x}{a-x}\right| + \frac{1}{2a^2}\cdot\frac{a}{a^2-x^2} + C \end{aligned}

Answer

I = \frac{1}{4a^3}\ln\left|\frac{a+x}{a-x}\right| + \frac{x}{2a^2(a^2-x^2)} + C

Example 3.3: Irreducible Factor Problem

Problem: Find $\int \frac{1}{(a^2+x^2)^2}\,dx$

Solution:

Method: Trigonometric Substitution

While this is not a typical partial fractions problem, it demonstrates how to handle irreducible factors.

Step 1: Trigonometric Substitution

Let $x = a\tan\theta$ , then $dx = a\sec^2\theta\,d\theta$

a^2 + x^2 = a^2(1 + \tan^2\theta) = a^2\sec^2\theta

Step 2: Substitute into Integral

\begin{aligned} I &= \int \frac{a\sec^2\theta}{a^4\sec^4\theta}\,d\theta = \frac{1}{a^3}\int \frac{1}{\sec^2\theta}\,d\theta \\ &= \frac{1}{a^3}\int \cos^2\theta\,d\theta = \frac{1}{a^3}\int \frac{1+\cos 2\theta}{2}\,d\theta \\ &= \frac{1}{2a^3}\left(\theta + \frac{\sin 2\theta}{2}\right) + C \end{aligned}

Step 3: Back-Substitute

From $\tan\theta = \frac{x}{a}$ , we get:

\begin{aligned} \theta &= \arctan\frac{x}{a} \\ \sin\theta &= \frac{x}{\sqrt{a^2+x^2}}, \quad \cos\theta = \frac{a}{\sqrt{a^2+x^2}} \\ \sin 2\theta &= 2\sin\theta\cos\theta = \frac{2ax}{a^2+x^2} \end{aligned}

I = \frac{1}{2a^3}\left(\arctan\frac{x}{a} + \frac{ax}{a^2+x^2}\right) + C

Answer

I = \frac{1}{2a^3}\arctan\frac{x}{a} + \frac{x}{2a^2(a^2+x^2)} + C

Example 3.4: Complex Partial Fractions

Problem: Find $\int \frac{3x+6}{(x-1)^2(x^2+x+1)}\,dx$

Solution:

Step 1: Check Denominator Factors

The denominator is already factored as $(x-1)^2(x^2+x+1)$ , where $x^2+x+1$ is irreducible ( $\Delta = 1-4 = -3 < 0$ )

Step 2: Set Up Partial Fractions

\frac{3x+6}{(x-1)^2(x^2+x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+x+1}

Step 3: Find Coefficients by Combining

After combining:

3x+6 = A(x-1)(x^2+x+1) + B(x^2+x+1) + (Cx+D)(x-1)^2

Expand and simplify:

\begin{aligned} A(x-1)(x^2+x+1) + B(x^2+x+1) + (Cx+D)(x-1)^2 &= Ax^3 + Bx^2 + Bx + B \\ &\quad + Cx^3 - 2Cx^2 + Cx + Dx^2 - 2Dx + D \\ &= (A+C)x^3 + (B-2C+D)x^2 \\ &\quad + (B+C-2D)x + (B+D) \end{aligned}

Compare coefficients:

\begin{cases} A + C = 0 \\ B - 2C + D = 0 \\ B + C - 2D = 3 \\ B + D = 6 \end{cases}

Solving: $A = -2, B = -3, C = 2, D = -1$

Step 4: Integrate

\begin{aligned} I &= -2\int\frac{1}{x-1}dx - 3\int\frac{1}{(x-1)^2}dx + \int\frac{2x-1}{x^2+x+1}dx \\ &= -2\ln|x-1| + \frac{3}{x-1} + \int\frac{2x+1-2}{x^2+x+1}dx \\ &= -2\ln|x-1| + \frac{3}{x-1} + \ln|x^2+x+1| - 2\int\frac{1}{x^2+x+1}dx \end{aligned}

For the last term, complete the square: $x^2+x+1 = (x+\frac{1}{2})^2 + \frac{3}{4}$

\int\frac{1}{x^2+x+1}dx = \frac{2}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}} + C

Answer

I = -2\ln|x-1| + \ln|x^2+x+1| - x + \frac{1}{3(x-1)} - \frac{4\sqrt{3}}{3}\arctan\frac{2x+1}{\sqrt{3}} + C

Example 3.5: Special Technique: Determining Coefficients

Problem: Find $\int \frac{x^2+ax-2}{(x-1)(x^2+1)}\,dx$ , where coefficient $a$ is to be determined so the result contains no arctangent function

Solution:

Step 1: Partial Fraction Decomposition

\frac{x^2+ax-2}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}

Step 2: Find Coefficient Relations

Combining:

x^2+ax-2 = A(x^2+1) + (Bx+C)(x-1)

Expand and compare coefficients:

\begin{cases} A + B = 1 \\ -B + C = a \\ A - C = -2 \end{cases}

Solving: $A = \frac{a-1}{2}, B = \frac{3-a}{2}, C = \frac{a+3}{2}$

Step 3: Analyze Integral Form

The integral is:

I = A\ln|x-1| + \int\frac{Bx+C}{x^2+1}dx = A\ln|x-1| + \frac{B}{2}\ln(x^2+1) + C\arctan x + K

To ensure the result contains no arctangent, we need $C = 0$ , that is:

\frac{a+3}{2} = 0 \quad \Rightarrow \quad a = -3

Step 4: Substitute and Calculate

When $a = -3$ , $A = -2, B = 3, C = 0$

I = -2\ln|x-1| + \frac{3}{2}\ln(x^2+1) + C = \ln\frac{(x^2+1)^{3/2}}{(x-1)^2} + C

Answer

When $a = -3$ :

I = \frac{3}{2}\ln(x^2+1) - 2\ln|x-1| + C

4. Key Techniques and Notes

Techniques for Finding Coefficients

1. Substitution Method

Substitute special values that make certain factors zero to quickly find some coefficients

2. Coefficient Comparison

After expansion, compare coefficients of like terms to establish a system of equations

3. Use of Symmetry

When the fraction has symmetry, you can reduce the number of undetermined coefficients

Common Pitfalls

1. Missing Repeated Factor Terms

For $(x-a)^n$ , write all terms from power 1 to n

2. Incorrect Irreducibility Check

A quadratic $ax^2+bx+c$ is irreducible only if $\Delta < 0$

3. Missing Integration Constant

Always add the integration constant $C$ in indefinite integrals