Integrate $\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$ 
Integrate $$\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$$.

My Try:
$$\int \frac{x\ln (x)}{x^3(1-1/x^2)^{3/2}} dx=\int \frac{\frac{1}{x^2}\ln (x)}{(1-\frac{1}{x^2})^{3/2}} dx.$$
Putting $x=\frac{1}{t}$,
I get
$$\int \frac{t^2\ln (\frac{1}{t})}{(1-t^2)^{3/2}(-t^2)} dt=\int \frac{\ln ({t})}{(1-t^2)^{3/2}} dt.$$
After this what to do?

Using the substitution $x=\sec(\theta)$, we get
$$
\begin{align}
\int\frac{x\log(x)}{\left(x^2-1\right)^{3/2}}\,\mathrm{d}x
&=-\int\log(x)\,\mathrm{d}\left(x^2-1\right)^{-1/2}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\frac{\mathrm{d}x}{x\sqrt{x^2-1}}\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\int\mathrm{d}\theta\\
&=-\frac{\log(x)}{\sqrt{x^2-1}}+\sec^{-1}(x)+C
\end{align}
$$
