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

$\tan 1-\frac{π}{4}$

We have $f(x)=\int\frac{(x^2+\sin^2x)}{1+x^2}\sec^2x\,dx$

$=\int\frac{x^2+(1-\cos^2x)}{1+x^2}.\sec^2x\,dx=\int(\sec^2x-\frac{1}{1+x^2})dx$

Thus, $f(x)=\tan x-\tan^{-1}x+C$

As $f(0)=0,C=0⇒f(x)=\tan x-\tan^{-1}x$

Hence, $f(1)=\tan 1-tan^{-1}1=\tan 1-\frac{π}{4}$