Evaluating the integral $\int (x^2-1)^{\frac{-3}{2}}dx$ I need to solve the differential equation:
$$\displaystyle f'(x) = (x^2-1)^{-\frac{3}{2}}, f(2)=\frac{3-2\sqrt 3}{3}$$
Which basically amounts to solving the integral $\int (x^2-1)^{-\frac{3}{2}} \mathrm{d} x$. I was thinkng of using $\int (x^2-1)^{-\frac{1}{2}} \mathrm{d} x = \operatorname{arcosh} x$ but I can't haven't managed to simplify it. Any ideas?
 A: You may write
$$
x=\cosh u, \quad dx=\sinh u\: du,
$$ giving
$$
\int\frac1{(x^2-1)^{3/2}}dx=\int\frac1{\sinh^2 u}du=-\text{cotanh}(u)+C= -\frac{x}{\sqrt{x^2-1}}+C
$$
A: I thought it might be instructive to present an approach that uses integration by parts, rather than hyperbolic trigonometric substitution.  To that end, we begin by writing
$$\begin{align}
\frac{1}{(x^2-1)^{3/2}}&=\frac{(1-x^2)+x^2}{(x^2-1)^{3/2}}\\\\
&=\frac{x^2}{(x^2-1)^{3/2}}-\frac{1}{\sqrt{x^2-1}}
\end{align} \tag 1$$
Using $(1)$ permits our writing 
$$\int \frac{1}{(x^2-1)^{3/2}}\,dx=\int \frac{x^2}{(x^2-1)^{3/2}}\,dx-\int \frac{1}{\sqrt{x^2-1}} \,dx \tag 2 $$
Integrating by parts the first integral on the right-hand side of $(2)$ yields
$$\int \frac{x^2}{(x^2-1)^{3/2}}\,dx=-x\frac{1}{\sqrt{x^2-1}}+\int \frac{1}{\sqrt{x^2-1}}\,dx +C \tag 3$$
Finally, using $(3)$ in $(2)$ reveals 
$$\int \frac{1}{(x^2-1)^{3/2}}\,dx=-x\frac{1}{\sqrt{x^2-1}}+C$$
which agrees with the result reported by @OlivierOloa from using hyperbolic trigonometric substitution!
