Partial Fractions

Overview

Partial fraction decomposition breaks a rational function into simpler fractions that are easier to integrate.

When to Use

For integrals of the form $\int \frac{P(x)}{Q(x)} \, dx$ where:

• $P$ and $Q$ are polynomials
• $\deg(P) < \deg(Q)$

If $\deg(P) \geq \deg(Q)$, divide first.

Cases

Case 1: Distinct Linear Factors

If $Q(x) = (a_1x + b_1)(a_2x + b_2)\cdots(a_nx + b_n)$:

$\frac{P(x)}{Q(x)} = \frac{A_1}{a_1x + b_1} + \frac{A_2}{a_2x + b_2} + \cdots + \frac{A_n}{a_nx + b_n}$

Case 2: Repeated Linear Factors

If $Q(x)$ contains $(ax + b)^n$:

$\cdots + \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n} + \cdots$

Case 3: Distinct Irreducible Quadratic Factors

If $Q(x)$ contains $(ax^2 + bx + c)$ where $b^2 - 4ac < 0$:

$\cdots + \frac{Ax + B}{ax^2 + bx + c} + \cdots$

Case 4: Repeated Irreducible Quadratic Factors

If $Q(x)$ contains $(ax^2 + bx + c)^n$:

$\cdots + \frac{A_1x + B_1}{ax^2 + bx + c} + \cdots + \frac{A_nx + B_n}{(ax^2 + bx + c)^n} + \cdots$

Finding Coefficients

Method 1: Equating Coefficients

Multiply both sides by $Q(x)$ and match coefficients of like powers.

Method 2: Strategic Substitution

Substitute values of $x$ that make factors zero.

Method 3: Cover-up Method (for distinct linear factors)

Cover up the factor in the denominator and evaluate at its root.

Examples

Example 1: Distinct linear factors

$\int \frac{1}{(x-1)(x+2)} \, dx$

Decompose: $\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$

$1 = A(x+2) + B(x-1)$

$x = 1$: $1 = 3A$, $A = \frac{1}{3}$

$x = -2$: $1 = -3B$, $B = -\frac{1}{3}$

$\int \left[\frac{1}{3(x-1)} - \frac{1}{3(x+2)}\right] dx = \frac{1}{3}\ln|x-1| - \frac{1}{3}\ln|x+2| + C$

$= \frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + C$

Example 2: Repeated linear factor

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

Decompose: $\frac{2x+3}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$

$2x + 3 = A(x-1) + B$

$x = 1$: $5 = B$

Coefficient of $x$: $2 = A$

$\int \left[\frac{2}{x-1} + \frac{5}{(x-1)^2}\right] dx = 2\ln|x-1| - \frac{5}{x-1} + C$

Example 3: Irreducible quadratic

$\int \frac{x+1}{x^2+1} \, dx$

$= \int \frac{x}{x^2+1} \, dx + \int \frac{1}{x^2+1} \, dx$

$= \frac{1}{2}\ln(x^2+1) + \arctan(x) + C$

Example 4: Linear and quadratic factors

$\int \frac{1}{(x-1)(x^2+1)} \, dx$

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

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

$x = 1$: $1 = 2A$, $A = \frac{1}{2}$

Expanding: $1 = Ax^2 + A + Bx^2 - Bx + Cx - C$

Comparing coefficients:

$x^2$: $0 = A + B$, so $B = -\frac{1}{2}$

$x^0$: $1 = A - C$, so $C = -\frac{1}{2}$

$\int \left[\frac{1}{2(x-1)} + \frac{-x/2 - 1/2}{x^2+1}\right] dx$

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

Example 5: Long division first

$\int \frac{x^3}{x^2-1} \, dx$

Divide: $x^3 \div (x^2-1) = x + \frac{x}{x^2-1}$

$= \int x \, dx + \int \frac{x}{(x-1)(x+1)} \, dx$

For second integral: $\frac{x}{(x-1)(x+1)} = \frac{1/2}{x-1} + \frac{1/2}{x+1}$

$= \frac{x^2}{2} + \frac{1}{2}\ln|x-1| + \frac{1}{2}\ln|x+1| + C$

$= \frac{x^2}{2} + \frac{1}{2}\ln|x^2-1| + C$