The value of the integral $\int\limits_{-1}^1 \log \left(x+\sqrt{x^2+1}\right) d x$, is

0

Let $f(x)=\log \left(x+\sqrt{x^2+1}\right)$. Then,

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

$\Rightarrow f(x)+f(-x)=\log \left(x^2+1-x^2\right)=\log 1=0 $

$\Rightarrow f(-x)=-f(x)$ for all $x \in(-1,1)$

$\Rightarrow f(x)$ is an odd function defined on $(-1,1)$

$\Rightarrow \int\limits_{-1}^1 f(x) d x=0$