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

Question

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.
Please help guys :)

Answer 1

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

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

Answer 2

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