<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] 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}
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