Question

Evaluate $\int \frac{1}{(x^2+1)(x^2+4)}dx$

Answer 1

Let $I=\int \frac{1}{(x^2+1)(x^2+4)}dx=\frac{1}{3}\int \left[\frac{(x^2+4)-(x^2+1)}{(x^2+4)(x^2+1)}\right]dx$

$I=\frac{1}{3}\int \frac{1}{x^2+1}-\frac{1}{x^2+4}dx=\frac{1}{3}{\tan }^{-1}x-\frac{1}{6}{\tan }^{-1}\frac{x}{2}+C$