Evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ How do I evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ ?
Note that this is a Q&A post and I've presented my solution below.
 A: To evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$
$\Rightarrow\int \frac{-2x}{(x^2-1)^{\frac{3}{2}}\sqrt{x^2+1}}dx$
Now, one substitution that works but is rather tricky to find is,
Let $u=\sqrt{\frac{x^2+1}{x^2-1}}$
$\Rightarrow \frac{1}{2}(\frac{x^2-1}{x^2+1})^{\frac{1}{2}} \frac{(x^2-1)2x-(x^2+1)2x}{(x^2-1)^2}dx=du$
$\Rightarrow (\frac{x^2-1}{x^2+1})^{\frac{1}{2}} \frac{-2x}{(x^2-1)^2}dx=du$
$\Rightarrow \frac{-2x}{(x^2-1)^{\frac{3}{2}}\sqrt{x^2+1}}dx=du$
The integral reduces to
$\int du=u+c=\sqrt{\frac{x^2+1}{x^2-1}}+c$
Thus $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx=\sqrt{\frac{x^2+1}{x^2-1}}+c$
A: Letting $x^2=\sec \theta$ transform the integral into
\begin{aligned}
I & =\int \frac{\sec \theta \tan \theta d \theta}{(1-\sec \theta) \tan \theta} \\
& =\int \frac{1}{\cos \theta-1} d \theta \\
& =\int \frac{\cos \theta+1}{-\sin ^2 \theta} d \theta \\
& =-\int \cot \theta \csc \theta d \theta-\int \csc ^2 \theta d \theta \\
& =\csc \theta+\cot \theta+C \\
& =\frac{x^2+1}{\sqrt{x^4-1}}+C
\end{aligned}
A: Substitute $x^2=\cosh 2t$ to integrate
\begin{align}
\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx
=-\int \text{csch}^2 t \ dt =\coth t +C
\end{align}
A: Another way to evaluate the OP’s integral is using Euler’s substitutions.
By letting $\;t=x^2\,,\;$ we get that
$\displaystyle\int\frac{2x}{\left(1-x^2\right)\sqrt{x^4-1}}\,\mathrm dx=\int\frac1{(1-t)\sqrt{t^2-1}}\,\mathrm dt\;.$
Now, we will apply Euler’s first substitution that is $\,\sqrt{t^2-1}=-t+u\,$ and obtain that
$\displaystyle\int\frac1{(1-t)\sqrt{t^2-1}}\,\mathrm dt=\int\frac{-2}{(u-1)^2}\,\mathrm du=\frac2{u-1}+C\;.$
Hence ,
$\displaystyle\int\frac{2x}{\left(1-x^2\right)\sqrt{x^4-1}}\,\mathrm dx=\frac2{x^2-1+\sqrt{x^4-1}}+C$

Addendum:
Note that
$\begin{align}\dfrac2{x^2-1+\sqrt{x^4-1}}+C&= \dfrac2{x^2-1+\sqrt{x^4-1}}+1+C-1=\\&=\dfrac{x^2+1+\sqrt{x^4-1}}{x^2-1+\sqrt{x^4-1}}+C-1=\\&=\dfrac{\sqrt{x^2+1}\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)}{\sqrt{x^2-1}\left(\sqrt{x^2-1}+\sqrt{x^2+1}\right)}+C-1=\\&=\dfrac{\sqrt{x^2+1}}{\sqrt{x^2-1}}+C-1=\\&=\sqrt{\dfrac{x^2+1}{x^2-1}}+C^*\,.\end{align}$
