Question

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

Answer 1

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

$\Rightarrow 1=A(x-3)+B(x+3)$

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 $B=\frac{1}{6}$

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

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

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

$=\frac{1}{6}\log \left|\frac{(x-3)}{(x+3)}\right|+C$