Let $f$ be a continuous function satisfying

 $f(x+y)=f\left(x\right)+f\left(y\right),$for each $x,y\in \mathbf{R}$

and $f\left(1\right)=2$then $\int \frac{f\left(x\right){\tan }^{-1}x}{{\left(1+x^2\right)}^2}\mathrm{d}x$ is equal to 

First show that $f\left(x\right)=2x$. Now , 

$I=\int \frac{f\left(x\right){\tan }^{-1}x}{{\left(1+x^2\right)}^2}dx=\int \frac{2x}{{\left(1+x^2\right)}^2}{\tan }^{-1}xdx$ 

Put $x=\tan \theta$ , so that

$\begin{array}{l}I=\int \frac{2\tan \theta }{{\sec }^4\theta }\theta {\sec }^2\theta d\theta \\ =\int \theta \sin \left(2\theta \right)d\theta \\ =-\frac{1}{2}\theta \cos 2\theta +\frac{1}{4}\sin 2\theta +C\\ =C-\frac{1}{2}\theta \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }+\frac{1}{2}\frac{\tan \theta }{1+{\tan }^2\theta }\\ =C-\frac{1}{2}\frac{1-x^2}{1+x^2}{\tan }^{-1}x+\frac{1}{2}\frac{x}{1+x^2}\end{array}$

$=C-\frac{1}{1+x^2}{\tan }^{-1}x+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\frac{x}{1+x^2}$