Answer
$$\sqrt{x^2-1}\ln (x) - \sqrt{x^2-1}+ \sec^{-1}x +C$$
Work Step by Step
Given
$$ \int \frac{x \ln x}{\sqrt{x^{2}-1}} d x$$
Use integration by parts , let
\begin{aligned}
u&=\ln x\ \ \ \ \ \ \ &dv&= \frac{x}{\sqrt{x^2-1}}dx\\
du&=\frac{1}{x}dx\ \ \ \ \ \ &v&=\sqrt{x^2-1}
\end{aligned}
Then
\begin{aligned}
\int \frac{x \ln x}{\sqrt{x^{2}-1}} d x&= \sqrt{x^2-1}\ln (x) -\int \frac{\sqrt{x^2-1}}{x}dx
\end{aligned}
To evaluate $\int \frac{\sqrt{x^2-1}}{x}dx$,
Let $$ x= \sec \theta \ \ \ \ \ \ dx= \sec\theta \tan \theta d\theta $$
Then
\begin{aligned}
\int \frac{\sqrt{x^2-1}}{x}dx&=\int \frac{\sqrt{\sec^2\theta-1}}{\sec\theta } \sec\theta \tan \theta d\theta\\
&=\int \tan^2\theta d\theta \\
&=\int (\sec^2\theta-1) d\theta \\
&=\tan \theta -\theta +C\\
&= \sqrt{x^2-1}- \sec^{-1}x +C
\end{aligned}
Hence
\begin{aligned}
\int \frac{x \ln x}{\sqrt{x^{2}-1}} d x&= \sqrt{x^2-1}\ln (x) - \sqrt{x^2-1}+ \sec^{-1}x +C
\end{aligned}
