# Integral of $\ln^2(x^2-1)/x^4$

I need to solve the following indefinite integral:
$$\int \frac{\log^2(x^2-1)}{x^4}dx.$$ ($\log$ is the natural log)
It's a past paper question from my uni exam so I don't think the answer is as complicated as WolframAlpha gives.

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$+1$ or $-1$ in the log ? – Claude Leibovici Mar 29 '14 at 9:43
sorry, -1 in the log – Djordje Mar 29 '14 at 9:57
As you have logarithm squared, you probably have no choice but to go for $u=\log (x^2-1)$, or even the entire square. However, I don't think you can avoid $Li$ function, unless Wolfram forgot to simplify. – orion Mar 29 '14 at 10:11
Yep, it seems like I'm getting something that resembles what Wolfram gave me - and in the end I have to integrate $\int{\frac{log(x+1)}{x-1}}$ and $\int{\frac{log(x-1)}{x+1}}$ which I can't integrate and Wolfram gives those two integrals as $Li$ functions. – Djordje Mar 29 '14 at 11:17

Hint: Start by integrating by parts with $u=\log^2(x^2-1)$ and $dv=\frac{1}{x^4}dx$. Then,

$$\int\frac{\log^2(x^2-1)}{x^4}dx=-\frac{\log^2(x^2-1)}{3x^3}+\frac43\int\frac{\log(x^2-1)}{x^2(x^2-1)}dx.$$

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Indeed, that's how I went about solving it. The problem was the $Li$ function that's a part of the solution later which we haven't learnt but after checking with my lecturer it turns out to be a mistype in the original exam question. – Djordje Mar 29 '14 at 18:27

take $log(x^2-1)$=t
after substituting you get
$$\int \frac{t^2e^t}{(e^t+1)^{5/2}}dt$$
write it as $$\int \frac{t^2(e^t+1-1)}{(e^t+1)^{5/2}}dt.$$
you get two integrals
$$\int \frac{1}{2}t^2(e^t+1)^{-3/2}dt+\int \frac{1}{2}t^2(e^t+1)^{-5/2}dt$$
i think you can go from here by parts

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Tried going that route, it doesn't work because you can't integrate $(e^t+1)^{-3/2}$ so that has to be your $u$ which means you'll just keep increasing ppwers of those expressions. – Djordje Mar 29 '14 at 11:14