Question

Integrate the rational functions.
$\int \frac{1}{x^2-9}dx$

Answer 1

$\int \frac{1}{x^2-9}dx=\int \frac{1}{x^2-3^2}dx=\int \frac{1}{(x+3)(x-3)}dx$
Let $\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}$
$\Rightarrow 1=A(x-3)+B(x+3)\Rightarrow 1=x(A+B)+(-3A+3B)$
On equating the coefficients of x and constant term on both sides, we get
A+B=0 and -3A+3B=1
On solving, we get $A=-\frac{1}{6}andB=\frac{1}{6}$
$\therefore \int \frac{1}{(x+3)(x-3)}dx=\int \frac{-1}{6(x+3)}+\int \frac{1}{6(x-3)}dx=-\frac{1}{6}log|x+3|+\frac{1}{6}log|x-3|+C=\frac{1}{6}log\left|\frac{x-3}{x+3}\right|+C[\because logb-loga=log\frac{b}{a}]$