<s>[INST] 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.
 [/INST] 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 $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx=\sqrt{\frac{x^2+1}{x^2-1}}+c$
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