<s>[INST] 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?
 [/INST] First integrate by parts, then substitute $x=\sec{\theta}$ $\implies \mathrm{dx}=\tan{\theta}\sec{\theta}\,\mathrm{d}\theta$:
$$\begin{align}
\int\frac{x\,\ln{\left(x\right)}}{\sqrt{x^2-1}}\,\mathrm{d}x
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{x^2-1}}{x}\,\mathrm{d}x\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{\sec^2{\theta}-1}}{\sec{\theta}}\,\tan{\theta}\sec{\theta}\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\tan^2{\theta}\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\left(\sec^2{\theta}-1\right)\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\tan{\theta}+\theta+\color{grey}{constant}\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\sqrt{x^2-1}+\sec^{-1}{x}+\color{grey}{constant}\\
\end{align}$$
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