$f(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$
Let $x=\tan \theta$
$\Rightarrow \quad d x=\sec ^2 \theta d \theta=\left(1+x^2\right) d \theta$
$f(x)=\int \frac{x^2 d x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$
$=\int \frac{\tan ^2 \theta \sec ^2 \theta d \theta}{\sec ^2 \theta(1+\sec \theta)}=\int \frac{\tan ^2 \theta d \theta}{1+\sec \theta}$
$=\int \frac{\sin ^2 \theta d \theta}{\cos \theta(1+\cos \theta)}$
$=\int \frac{1-\cos ^2 \theta d \theta}{\cos \theta(1+\cos \theta)}$
$=\int \frac{(1-\cos \theta) d \theta}{\cos \theta}$
$=\int \sec \theta d \theta-\int d \theta$
$=\log \left(x+\sqrt{1+x^2}\right)-\tan ^{-1} x+C$
$\therefore \quad f(0)=\log (0+\sqrt{1+0})-\tan ^{-1}(0)+C$
$0=\log 1-0+C$
$\Rightarrow \quad C=0$
$\therefore \quad f(1)=\log \left(1+\sqrt{1+1^2}\right)-\tan ^{-1}(1)$
$=\log (1+\sqrt{2})-\frac{\pi}{4}$