The integral $\int \frac{x^2(1-\ln(x))}{(\ln(x))^4-x^4} dx$? $$\int \frac{x^2(1-\ln(x))}{\ln^4(x)-x^4} dx$$
I tried to factorize the denominator so that I could apply integration by parts but that didn't help me at all.
 A: HINT:
$$\mathcal{I}(x)=\int\frac{x^2\left(1-\ln(x)\right)}{\ln^4(x)-x^4}\space\text{d}x=$$
$$\frac{1}{2}\int\frac{\ln(x)-1}{x^2+\ln^2(x)}\space\text{d}x-\frac{1}{4}\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x+\frac{1}{4}\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x$$


*

*Now, for $\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x$,
substitute $u=\ln(x)-x$ and
$\text{d}u=\left(\frac{1}{x}-1\right)\space\text{d}x$:
$$\int\frac{1-x}{x\left(\ln(x)-x\right)}\space\text{d}x=\int\frac{1}{u}\space\text{d}u=\ln\left|u\right|+\text{C}=\ln\left|\ln(x)-x\right|+\text{C}$$

*Now, for $\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x$,
substitute $s=\ln(x)+x$ and
$\text{d}s=\left(\frac{1}{x}-1\right)\space\text{d}x$:
$$\int\frac{1+x}{x\left(x+\ln(x)\right)}\space\text{d}x=\int\frac{1}{s}\space\text{d}s=\ln\left|s\right|+\text{C}=\ln\left|\ln(x)+x\right|+\text{C}$$

A: $$I=\int \frac{x^2(1-\ln(x))}{\ln^4(x)-x^4} dx$$
$$I=\int\frac{\frac{1-\ln x}{x^2}}{\left(\frac{\ln x }{x}\right)^4-1}dx$$
Now put $\displaystyle \frac{\ln x}{x}=t,$ Then $\displaystyle \frac{1-\ln x}{x^2}dx=dt$
So $$$I=\int\frac{1}{t^4-1}dt = \int\frac{1}{(1-t^2)(1+t^2)}dt =\frac{1}{2}\int\left[\frac{1}{1-t^2}+\frac{1}{1+t^2}\right]dt$$
