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: 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 $x=\cosh t$.
A: First integrate by parts, then substitute $x=\sec{\theta}$ $\implies \mathrm{dx}=\tan{\theta}\sec{\theta}\,\mathrm{d}\theta$:
$$\begin{align}
\int\frac{x\,\ln{\left(x\right)}}{\sqrt{x^2-1}}\,\mathrm{d}x
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{x^2-1}}{x}\,\mathrm{d}x\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\frac{\sqrt{\sec^2{\theta}-1}}{\sec{\theta}}\,\tan{\theta}\sec{\theta}\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\tan^2{\theta}\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\int\left(\sec^2{\theta}-1\right)\,\mathrm{d}\theta\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\tan{\theta}+\theta+\color{grey}{constant}\\
&=\sqrt{x^2-1}\,\ln{\left(x\right)}-\sqrt{x^2-1}+\sec^{-1}{x}+\color{grey}{constant}\\
\end{align}$$
A: 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*}
A: $$
\begin{aligned}
\int \frac{x \ln x}{\sqrt{x^{2}-1}} d x &=\int \ln x d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{\sqrt{x^{2}-1}}{x} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x^{2}} d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int\left(1-\frac{1}{x^{2}}\right) d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\sqrt{x^{2}-1}+\int \frac{d \sqrt{x^{2}-1}}{\left(\sqrt{x^{2}-1}\right)^{2}+1}\\
&=\sqrt{x^{2}-1} \ln x-\sqrt{x^{2}-1}+\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
$$
