The value of $\int \frac{\log _e\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}} d x$, is

$\frac{1}{2}\log _e\left(x+\sqrt{x^2+1}\right)^2+C$

Let $\log_e(x+\sqrt{x^2+1})=t$

$⇒\frac{dt}{dx}=\frac{1}{x+\sqrt{x^2+1}}.\frac{d}{dx}(x+\sqrt{x^2+1})$

$=\frac{1}{x+\sqrt{x^2+1}}.\left(1+\frac{1.2x}{2\sqrt{x^2+1}}\right)$

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

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

$I=\int\log _e\left(x+\sqrt{x^2+1}\right).\frac{dx}{\sqrt{x^2+1}}=\int t\,dt$

$=\frac{t^2}{2}+C$

$=\frac{1}{2}.\left(\log _e\,x+\sqrt{x^2+1}\right)^2+C$