$\int \frac{\log \left(x+\sqrt{x^2+1}\right)}{\sqrt{1+x^2}} d x$ is equal to :

$\frac{1}{2}\left[\ln \left(x+\sqrt{x^2+1}\right)\right]^2+c$

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

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

∴  $\frac{1}{x+\sqrt{x^2+1}}\left[1+\frac{x}{\sqrt{x^2+1}}\right] d x=d t$

$\Rightarrow \frac{1}{\sqrt{x^2+1}} dx=dt$

$\Rightarrow I=\int t d t=\frac{t^2}{2}+c$

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

Hence (3) is the correct answer.