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?

You can evaluate the integral without using secant substitution. Here is the way. 
Since we know that
\begin{gather*}
 \mathrm{d}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}}\mathrm{d} x,
\end{gather*}
we can calculate, by differentiation by parts, as follows,
\begin{align*}
 &\quad \frac{x\ln(x)}{\sqrt{x^2-1}} \mathrm{d} x=\ln(x)\mathrm{d} \sqrt{x^2-1}=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{\sqrt{x^2-1}}{x}\mathrm{d} x\\
 &= \mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x^2-1}{x\sqrt{x^2-1}}\mathrm{d} x\\
 &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\frac{x}{\sqrt{x^2-1}}\mathrm{d} x+\frac{1}{x\sqrt{x^2-1}}\mathrm{d} x\\
 &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)+\frac{1}{x^2\sqrt{1-\left(\frac{1}{x}\right)^2}}\mathrm{d} x\\
 &=\mathrm{d} \big(\ln(x)\sqrt{x^2-1}\big)-\mathrm{d}\big(\sqrt{x^2-1}\big)-\frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^ 2}}\mathrm{d} \left(\frac{1}{x}\right)\\
 &=\mathrm{d}\left(\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)\right),
\end{align*}
hence we have
\begin{gather*}
 \int\frac{x\ln(x)}{\sqrt{x^2-1}}\mathrm{d} x=\ln(x)\sqrt{x^2-1}-\sqrt{x^2-1}-\arcsin\left(\frac{1}{x}\right)+C.
\end{gather*}
