Integration by Partial Fractions

By the end of this lesson, you will be able to integrate any rational function using partial fraction decomposition, handle all cases (distinct linear, repeated linear, irreducible quadratic), and complete the full integration.

The Big Idea

Every rational function $P(x)/Q(x)$ (with $\deg P < \deg Q$) can be decomposed into partial fractions and each piece integrated separately. This technique makes rational function integration completely systematic.

Setup (Review from Pre-Calculus)

1. Ensure proper ($\deg P < \deg Q$); otherwise divide first
2. Factor $Q(x)$ completely over the reals
3. Write the partial fraction form based on the factors
4. Solve for the constants
5. Integrate each piece

Integration of Each Piece

Form	Integral
$\dfrac{A}{x-a}$	$A\ln
$\dfrac{A}{(x-a)^n}$ ($n \geq 2$)	$\dfrac{A}{-(n-1)(x-a)^{n-1}} + C$
$\dfrac{Ax+B}{x^2+bx+c}$	Complete the square, then $\arctan$

Example 1 — Distinct Linear Factors

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

Decompose: $\dfrac{A}{x-3}+\dfrac{B}{x+2}$. Solve: $A = 1, B = 1$.

$= \int\frac{1}{x-3},dx + \int\frac{1}{x+2},dx = \ln|x-3| + \ln|x+2| + C$

$= \ln|(x-3)(x+2)| + C$

Example 2 — Repeated Linear Factor

$$\int\frac{5x-3}{x^2(x-1)},dx$$

Form: $\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x-1}$

Solve: $5x-3 = Ax(x-1)+B(x-1)+Cx^2$

$x=0$: $-3 = -B \implies B = 3$

$x=1$: $2 = C$

$x^2$ coefficient: $0 = A+C \implies A = -2$

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

Example 3 — Irreducible Quadratic

$$\int\frac{x^2+4}{x^3-x^2+x-1},dx$$

Factor: $x^3-x^2+x-1 = x^2(x-1)+(x-1) = (x-1)(x^2+1)$

Form: $\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}$

$x^2+4 = A(x^2+1)+(Bx+C)(x-1)$

$x=1$: $5 = 2A \implies A = 5/2$

$x^2$: $1 = A+B \implies B = -3/2$

$x^0$: $4 = A-C \implies C = 1/2$

$= \frac{5}{2}\ln|x-1| - \frac{3}{4}\ln(x^2+1) + \frac{1}{2}\tan^{-1}x + C$