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

$\tan 1+\frac{\pi}{4}$

We have,

$f(x)=\int\left\{\frac{\left(x^2+1\right)+\left(1-\sin ^2 x\right)}{1+x^2}\right\} \sec ^2 x d x$

$\Rightarrow f(x)=\int\left(\sec ^2 x+\frac{1}{1+x^2}\right) d x$

$\Rightarrow f(x)=\tan x+\tan ^{-1} x+C$

∴  $f(0)=0 \Rightarrow C=0$

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

$\Rightarrow f(1)=\tan 1+\frac{\pi}{4}$