Question

Solve:
$\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}dx$

Answer 1

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

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

$\Rightarrow \frac{1}{3}\int \frac{dx}{\left(x{z}^{-}4\right)}-\frac{1}{3}\int \frac{dx}{x^2+2^2}$

$\Rightarrow \frac{1}{3}[{\tan }^{2}x-\frac{1}{2}-{\tan }^{-1}\frac{x}{2}]+c$

$I\Rightarrow \frac{1}{6}[2{\tan }^{2}x-{\tan }^{-1}x/2]+c$