The value of $\int \frac{1}{x+\sqrt{x-1}} d x$, is

$\log (x+\sqrt{x-1})-\frac{2}{3} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C$

Let $x-1=t^2$. Then,

$\int \frac{1}{x+\sqrt{x-1}} d x=2 \int \frac{t}{t^2+t+1} d t=\int \frac{(2 t+1)-1}{t^2+t+1} d t$

$=\int \frac{2 t+1}{t^2+t+1} d t-\int \frac{1}{\left(t+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d t$

$=\log \left(t^2+t+1\right)-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t+1}{\sqrt{3}}\right)+C$

$=\log (x+\sqrt{x-1})-\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C$