Question

Integrate the rational function: $\frac{1}{x^2-9}$

Answer 1

Let $\frac{1}{(x+3)(x-3)} = \frac{A}{(x+3)}+\frac{B}{(x-3)}$

1 = A (x-3)+B(x+3)
Equating the coefficients of x and constant term, we obtain
A + B = 0
−3A + 3B = 1
On solving, we obtain

A =- $\frac{1}{6}$and B = $\frac{1}{6}$

∴ $\frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}$

$\Rightarrow \int\frac{1}{(xx^2-9)}dx=\int\bigg(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\bigg)dx$

= $-\frac{1}{6}\log\mid x+3 \mid +\frac{1}{6}\log \mid x-3 \mid+C$

$\frac{1}{6}\log \mid\frac{(x-3)}{(x+3)} \mid+C$