$$
\begin{aligned}
& \text { Let } f(x)=\int\left(\frac{x^2+\sin ^2 x}{1+x^2}\right) \sec ^2 x d x \\
& =\int \frac{x^2 \sec ^2 x+\frac{\sin ^2 x}{\cos ^2 x}}{1+x^2} d x
\end{aligned}
$$

$$
\begin{aligned}
& =\int \frac{x^2 \sec ^2 x+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2\left(1+\tan ^2 x\right)+\tan ^2 x}{1+x^2} d x \\
& =\int \frac{x^2+\tan ^2 x\left(1+x^2\right)}{1+x^2} d x \\
& =\int \frac{x^2}{1+x^2} d x+\int \tan ^2 x d x \\
& =\int \frac{x^2+1-1}{1+x^2} \mathrm{dx}+\int\left(\sec ^2 \mathrm{x}-1 \mathrm{dx}\right. \\
& =\int 1 d x-\int \frac{d x}{1+x^2}+\int \sec ^2 x d x-\int d x \\
& =-\tan -1 x+\tan ^{-1} x+\mathrm{c} \\
& \text { Given: } f(0)=0 \\
& \Rightarrow f(0)=-\tan ^{-1} 0+\tan ^2+c \\
& \Rightarrow c=0 \\
& \therefore f(x)=-\tan ^{-1} x+\tan ^2 x
\end{aligned}
$$
Now,
$$
f(1)=-\tan ^{-1}(1)+\tan 1=\tan 1-\frac{\pi}{4}
$$