Question

If $\int \frac{dx}{(x^2+1)(x^2+4)}=k{\tan }^{-1}x+l{\tan }^{-1}\frac{x}{2}+C$, then

• A. $k=\frac{1}{3}$
• B. $l=\frac{2}{3}$
• C. $k=-\frac{1}{3}$
• D. $l=-\frac{1}{6}$

Answer 1

The correct option is C $k=-\frac{1}{3}$

Given, $\int \frac{dx}{(x^2+1)(x^2+4)}=k{\tan }^{-1}x+l{\tan }^{-1}\frac{x}{2}+C$
Differentiating w.r.t. $x$ , we get
$\frac{1}{(x^2+1)(x^2+4)}=\frac{k}{(x^2+1)}+\frac{l}{{\left(\frac{x}{2}\right)}^2+1}$
$\frac{1}{(x^2+1)(x^2+4)}=\frac{k}{(x^2+1)}+\frac{4l}{(x^2+4)}$
$\Rightarrow 1=(k+4l)x^2+4k+4l$
On comparing, we get
$k+4l=0,4k+4l=1$
$\Rightarrow k=-\frac{1}{3}$ and $l=-\frac{1}{12}$