In partial fraction decomposition you split a rational expression into simpler fractions. Here is the payoff: each simple fraction has an antiderivative you already know, from integrals giving logarithms and arctangents.
Por qué funciona
Adding fractions merges denominators: $\frac{2}{x - 1} + \frac{3}{x + 1} = \frac{5x - 1}{(x - 1)(x + 1)}$. Partial fractions runs that backward. It works because each simple piece is easy to integrate:
$$\begin{aligned} &\int \frac{A}{x - a}dx \\ &= A\ln|x - a|, \quad \int \frac{B}{(x - a)^{2}}dx \\ &= -\frac{B}{x - a}, \quad \int \frac{dx}{x^{2} + k^{2}} \\ &= \frac{1}{k}\arctan\frac{x}{k}. \end{aligned}$$
The decomposition only makes sense for a proper fraction (top degree less than bottom degree), because a sum of proper fractions is always proper.
Idea clave
(1) If the top's degree is at least the bottom's, divide first. (2) Factor the bottom. (3) Write one term per factor: $\frac{A}{x - a}$ for each linear factor, $\frac{A}{x - a} + \frac{B}{(x - a)^{2}}$ for a squared factor, $\frac{Bx + D}{x^{2} + bx + c}$ for an irreducible quadratic. (4) Find the constants. (5) Integrate term by term.
Ejemplo resuelto: Two linear factors
Find $\int \frac{5x - 1}{(x - 1)(x + 1)}\,dx$.
- Write $\frac{5x - 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$, so $5x - 1 = A(x + 1) + B(x - 1)$.
- $x = 1$: $4 = 2A$, so $A = 2$. $x = -1$: $-6 = -2B$, so $B = 3$.
- $$\begin{aligned} &\int \left(\frac{2}{x - 1} + \frac{3}{x + 1}\right)dx \\ &= 2\ln|x - 1| + 3\ln|x + 1| + C. \end{aligned}$$
Respuesta: $2\ln|x - 1| + 3\ln|x + 1| + C$
Error común
Splitting before dividing. For $\frac{x^{2} + 1}{x^{2} - 1}$, trying $\frac{A}{x - 1} + \frac{B}{x + 1}$ cannot work: that sum always has a top of lower degree than the bottom, but this fraction does not. Check the degrees first. Here $\frac{x^{2} + 1}{x^{2} - 1} = 1 + \frac{2}{x^{2} - 1}$, and only the remainder gets split.
Ejemplo resuelto: Divide first
Find $\int \frac{x^{2} + 1}{x^{2} - 1}\,dx$.
- Same degree on top and bottom: $\frac{x^{2} + 1}{x^{2} - 1} = 1 + \frac{2}{x^{2} - 1}$.
- $\frac{2}{(x - 1)(x + 1)} = \frac{1}{x - 1} - \frac{1}{x + 1}$ (cover-up: $A = \frac{2}{2}$, $B = \frac{2}{-2}$).
- $$\begin{aligned} &\int \left(1 + \frac{1}{x - 1} - \frac{1}{x + 1}\right)dx \\ &= x + \ln|x - 1| - \ln|x + 1| + C. \end{aligned}$$
Respuesta: $x + \ln\left|\frac{x - 1}{x + 1}\right| + C$
Ejemplo resuelto: A repeated factor
Find $\int \frac{x + 3}{x(x + 1)^{2}}\,dx$.
- $\frac{x + 3}{x(x + 1)^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{D}{(x + 1)^{2}}$, so $x + 3 = A(x + 1)^{2} + Bx(x + 1) + Dx$.
- $x = 0$: $A = 3$. $x = -1$: $2 = -D$, so $D = -2$. The $x^{2}$ terms: $0 = A + B$, so $B = -3$.
- $$\begin{aligned} &\int \left(\frac{3}{x} - \frac{3}{x + 1} - \frac{2}{(x + 1)^{2}}\right)dx \\ &= 3\ln|x| - 3\ln|x + 1| + \frac{2}{x + 1} + C. \end{aligned}$$
Respuesta: $3\ln|x| - 3\ln|x + 1| + \frac{2}{x + 1} + C$
Ejemplo resuelto: An irreducible quadratic
Find $\int \frac{2x^{2} + 2x + 3}{x(x^{2} + 1)}\,dx$.
- $\frac{2x^{2} + 2x + 3}{x(x^{2} + 1)} = \frac{A}{x} + \frac{Bx + D}{x^{2} + 1}$, so $$\begin{aligned} &2x^{2} + 2x + 3 \\ &= A(x^{2} + 1) + (Bx + D)x. \end{aligned}$$
- $x = 0$: $A = 3$. Then $x^{2}$: $2 = A + B$, so $B = -1$; $x$: $D = 2$.
- $\int \frac{-x}{x^{2} + 1}dx = -\frac{1}{2}\ln(x^{2} + 1)$ and $\int \frac{2}{x^{2} + 1}dx = 2\arctan x$.
- Answer: $$\begin{aligned} &3\ln|x| - \frac{1}{2}\ln(x^{2} + 1) \\ &\quad + 2\arctan x + C. \end{aligned}$$
Respuesta: $3\ln|x| - \frac{1}{2}\ln(x^{2} + 1) + 2\arctan x + C$
We named the constants $A$, $B$, $D$ to keep $C$ free for the constant of integration.