If $f\left(x\right)=\int \left(\frac{x^2+{\sin }^{2}x}{1+x^2}\right){\sec }^{2}xdx$ and $f\left(0\right)=0$, then f(1) equals:

Solution:

We have,

$\begin{array}{l}f\left(x\right)=\int \left\{\frac{\left(x^2+1\right)-\left(1-{\sin }^{2}x\right)}{1+x^2}\right\}{\sec }^{2}xdx\\ \Rightarrow f\left(x\right)=\int \left({\sec }^{2}x-\frac{1}{1+x^2}\right)dx\\ \Rightarrow f\left(x\right)=\tan x-{\tan }^{-1}x+C\\ \therefore f\left(0\right)=0\Rightarrow C=0\end{array}$

Hence, $f\left(x\right)=\tan x-{\tan }^{-1}x$

$\Rightarrow f\left(1\right)=\tan 1-\frac{\pi }{4}$