<s>[INST] Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$ Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$

My Try:
$\int\frac{3x^2+1}{(x^2-1)^3}dx$
Put $x=\sec\theta$
$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$
Converting to $\sin$ and $\cos$ we get
$=\int\frac{(3+\cos^2\theta)\cos^2\theta}{\sin^5\theta}d\theta$
I could not solve it further.
 [/INST] Another way is, simply, to take the derivative of $$\frac{-x}{(x^2-1)^2}+c$$ and verify that this derivative is no other that $$\frac{3x^2+1}{(x^2-1)^3}$$.
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