TITLE: It's an indefinite integral $\int\frac{x^2(1- \ln(x))}{(\ln(x))^4-x^4}\,dx.$
QUESTION [5 upvotes]: $$\int\frac{x^2(1- \ln(x))}{(\ln(x))^4-x^4}\,dx.$$

I try to solve it by factoring, but got stuck in the problem. Please help anyone.

REPLY [8 votes]: Hint:
After cheating with WA, here is the solution: notice that "by magic",
$$\int\frac{x^2(1-\log x)}{(\log x)^4-x^4}dx
=\int\frac{\dfrac{1-\log x}{x^2}}{\dfrac{(\log x)^4}{x^4}-1}dx
=\int\frac1{\dfrac{(\log x)^4}{x^4}-1}d\frac{\log x}x.$$
Now
$$\frac1{u^4-1}=\frac1{4 (u-1)} -\frac 1{4 (1 + u)} - \frac1{2 (1 + u^2)}$$ which is easy to integrate.

REPLY [4 votes]: Let $u=\frac{\ln x }{x}$ then $du = \frac{1 - \ln x }{x^2} dx$ so $dx =\frac{x^2 du}{1-\ln x } $
$$\int \frac{x^2 (1-\ln x )}{x^4 (u^4-1)}\frac{x^2 du}{1-\ln x } $$