Answer
$x\ln(x+\sqrt{x^2-1})-\sqrt{x^2-1}+C$
Work Step by Step
Let
$I=\int\ln(x+\sqrt{x^2-1})\;dx$ ______(1)
$I=\int\ln(x+\sqrt{x^2-1})\cdot 1\;dx$
Using integration by parts with $f(x)=\ln(x+\sqrt{x^2-1})\;,g(x)=1$
$I=\ln(x+\sqrt{x^2-1})\int dx-\int\left(\frac{d}{dx}(\ln(x+\sqrt{x^2-1})\int dx\right)dx$
$I=x\ln(x+\sqrt{x^2-1})-\int\left(\frac{1+\frac{2x}{2\sqrt{x^2-1}}}{x+\sqrt{x^2-1}}\right)xdx$
$I=x\ln(x+\sqrt{x^2-1})-\int\left(\frac{\frac{x+\sqrt{x^2-1} }{\sqrt{x^2-1}}}{x+\sqrt{x^2-1}}\right)x\;dx$
$I=x\ln(x+\sqrt{x^2-1})-\int\frac{x}{\sqrt{x^2-1}}\:dx$
Substitute $x^2-1=t$ _____(2)
$\;\;\;\;\;\;\;\;\; \Rightarrow 2xdx=dt$
$I=x\ln(x+\sqrt{x^2-1})-\frac{1}{2}\int\frac{dt}{\sqrt t}$
$I=x\ln(x+\sqrt{x^2-1})-\frac{1}{2}\int t^{-\frac{1}{2}}dt$
$I=x\ln(x+\sqrt{x^2-1})-\frac{1}{2}[2t^{\frac{1}{2}}]+C$
Where $C$ is constant of jntegration
From (2)
$I=x\ln(x+\sqrt{x^2-1})-\sqrt{x^2-1}+C$
Hence $\;I=x\ln(x+\sqrt{x^2-1})-\sqrt{x^2-1}+C$.
