If $\int \frac{\log (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=\left(gof\right)\left(x\right)+c then$

$I=\int \frac{\log (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$

$Put \log \left(x+\sqrt{1+x^2}\right)=t$

$\Rightarrow \frac{1}{x+\sqrt{1+x^2}}\left[x+\frac{1}{2\sqrt{1+x^2}}2x\right]dx=dt$

$I=\frac{{\left[\log (x+\sqrt{1+x^2})\right]}^2}{2}+C$

$$\begin{gathered} =\left(gof\right)\left(x\right)+C=g\left(f\left(x\right)\right)+C \\ where g\left(x\right)=\frac{x^2}{2} and f\left(x\right)=\log \left(x+\sqrt{1+x^2}\right) \end{gathered}$$