Question

Evaluate $\int \frac{1}{x^2-9}dx$.

Answer 1

Let $\frac{1}{x^2-9}=\frac{1}{(x+3)(x-3)}=\frac{A}{x+3}+\frac{B}{x-3}$

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

$(A+B)x+(3A-3B-1)=0$

By equating coefficient and on solving, we get

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

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

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

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