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

$f\left(x\right)=\int \frac{x^{2}dx}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$

$$\begin{gathered} Put x=\tan \theta \\ dx=se{c}^2\theta d\theta \end{gathered}$$

$$\begin{gathered} f\left(x\right)=\int \frac{{\tan }^2\theta .se{c}^2\theta }{\left(1+{\tan }^2\theta \right)\left(1+\sqrt{1+{\tan }^2\theta }\right)}d\theta \\ =\int \frac{{\tan }^2\theta }{\left(1+sec\theta \right)}d\theta \\ =\int \frac{{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}}{{\displaystyle \frac{\cos \theta +1}{\cos \theta }}}d\theta \\ =\int \frac{{\sin }^2\theta }{\cos \theta \left(\cos \theta +1\right)}d\theta \\ =\int \frac{1-{\cos }^2\theta }{\cos \theta \left(\cos \theta +1\right)}d\theta \\ =\int \left(\frac{1-\cos \theta }{\cos \theta }\right)d\theta \\ =\int \left(sec\theta -1\right)d\theta \\ =\log \left|sec\theta +\tan \theta \right|-\theta +C \\ =\log \left|\sqrt{1+x^2}+x\right|-{\tan }^{-1}x+C \\ f\left(0\right)=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow \log \left|\sqrt{1+0^2}+0\right|-{\tan }^{-1}0+C=0 \\ 0+C=0\Rightarrow C=0 \\ \therefore f\left(1\right)=\log \left|\sqrt{1+1^2}+1\right|-{\tan }^{-1}1 \\ =\log \left(1+\sqrt{2}\right)-\frac{\pi }{4} \end{gathered}$$