If \(\mathrm{f}(x)=\int \frac{x^2 \mathrm{~d} x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}\) and \(\mathrm{f}(0)=0\), then \(f(1)\) is

\(\text { Put } x=\tan \theta \Rightarrow \mathrm{d} x=\sec ^2 \theta \mathrm{d} \theta \)
\( \therefore \mathrm{f}(x) =\int \frac{\tan ^2 \theta \sec ^2 \theta \mathrm{d} \theta}{\sec ^2 \theta(1+\sec \theta)} \)
\( =\int \frac{\tan ^2 \theta \mathrm{d} \theta}{1+\sec \theta} \)
\( =\int \frac{\sin ^2 \theta \mathrm{d} \theta}{\cos \theta(1+\cos \theta)} \)
\( =\int \frac{1-\cos ^2 \theta \mathrm{d} \theta}{\cos \theta(1+\cos \theta)} \)
\( =\int \frac{(1-\cos \theta) \mathrm{d} \theta}{\cos \theta} \)
\( =\int \sec \theta \mathrm{d} \theta-\int \mathrm{d} \theta \)
\( =\log |\sec \theta+\tan \theta|-\theta+\mathrm{c} \)
\( \mathrm{f}(x) =\log \left|x+\sqrt{1+x^2}\right|-\tan ^{-1} x+\mathrm{c} \)
\( \therefore \mathrm{f}(0) =\log |0+\sqrt{1+0}|-\tan ^{-1}(0)+\mathrm{c} \)
\( \Rightarrow 0 =\log 1-0+\mathrm{c} \Rightarrow \mathrm{c}=0 \)
\( \therefore \mathrm{f}(x) =\log \left|x+\sqrt{1+x^2}\right|-\tan ^{-1} x \)
\( \therefore \mathrm{f}(1) =\log \left|1+\sqrt{1+1^2}\right|-\tan ^{-1}(1) \)
\( \therefore \log (1+\sqrt{2})-\frac{\pi}{4}\)