Solve $\int \frac{x\ln(x)}{\sqrt{x^2 - 1}}$ My Work
$x = \tan\theta$
$dx = \sec^2\theta d\theta$
$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$
$\int \sec\theta \tan\theta ln(tan\theta)$ 
$u = \ln(\tan\theta)$
$du = \frac{\sec^2 \theta}{\tan \theta}$
$dv = \sec \theta \tan \theta d\theta$
$v = \sec \theta$
$\sec\theta\ln\tan\theta - \int \frac{\sec^3\theta}{\tan \theta}$
$\sec\theta\ln\tan\theta - \int \sec^2\theta \csc\theta d\theta$
I'm stuck after here. Parts doesn't look particularly appealing. I don't see an easy substitution. Brain is pretty tired at this point. Anyone know what to do?

$$
\begin{aligned}
\int \frac{x \ln x}{\sqrt{x^{2}-1}} d x &=\int \ln x d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{\sqrt{x^{2}-1}}{x} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x^{2}} d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int\left(1-\frac{1}{x^{2}}\right) d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\sqrt{x^{2}-1}+\int \frac{d \sqrt{x^{2}-1}}{\left(\sqrt{x^{2}-1}\right)^{2}+1}\\
&=\sqrt{x^{2}-1} \ln x-\sqrt{x^{2}-1}+\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
$$
