Question

State whether True=1 or False=0
$\int \frac{x^2}{(x^2+1)(x^2+4)}dx=\frac{-1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$

Answer 1

$\int \frac{x^2}{(x^2+1)(x^2+4)}$
We will solve using partial fraction method
Let $x^2=y$
$\therefore \frac{x^2}{(x^2+1)(x^2+4)}=\frac{y}{(y+1)(y+4)}$
$\frac{y}{(y+1)(y+4)}=\frac{A}{(y+1)}+\frac{B}{(y+4)}$
$\Rightarrow y=A(y+4)+B(y+1)$
When $y=-1\Rightarrow A=\frac{-1}{3}$
When $y=-4\Rightarrow B=\frac{4}{3}$
$\Rightarrow \frac{y}{(y+1)(y+4)}=\frac{-1}{3(y+1)}+\frac{4}{3(y+4)}$
$\Rightarrow \frac{x^2}{(x^2+1)(x^2+4)}=\frac{-1}{3(x^2+1)}+\frac{4}{3(x^2+4)}$
Integrating both sides w.r.t. x , we get
$\int \frac{x^2}{(x^2+1)(x^2+4)}dx=\frac{-1}{3}{\tan }^{-1}x+\frac{2}{3}{\tan }^{-1}\left(\frac{x}{2}\right)+C$