# 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?

A start: Integrate by parts, letting $u=\ln x$ and $dv=\frac{x}{\sqrt{x^2-1}}\,dx$.
Remark: The substitution $x=\tan\theta$ is not good for $\sqrt{x^2-1}$. There the appropriate thing is $x=\sec\theta$ or $x=\cosh t$.
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}