∫ (x^2(1-ln x))/((ln x)^4-x^4) d x is equal to

Let $I=\int \frac{x^2(1-\ln x)}{(\ln x)^4-x^4} d x$. Then,

$I=\int \frac{x^2(1-\ln x)}{\left((\ln x)^2+x^2\right)(\ln x+x)(\ln x-x)} d x$

$\Rightarrow I=\int \frac{\frac{1-\ln x}{x^2}}{\left\{\left(\frac{\ln x}{x}\right)^2+1\right\}\left(\frac{\ln x}{x}+1\right)\left(\frac{\ln x}{x}-1\right)} d x$

$\Rightarrow I=\int \frac{1}{\left(t^2+1\right)(t+1)(t-1)} d t$, where $t=\frac{\ln x}{x}$

$\Rightarrow I=\int \frac{1}{\left(t^2+1\right)\left(t^2-1\right)} d t$

$\Rightarrow I=\frac{1}{2} \int\left(\frac{1}{t^2-1}-\frac{1}{t^2+1}\right) d t=\frac{1}{4} \ln \left|\frac{t-1}{t+1}\right|-\frac{1}{2} \tan ^{-1} t+C$

$\Rightarrow I=\frac{1}{4} \ln \left|\frac{\frac{\ln x}{x}-1}{\frac{\ln x}{x}+1}\right|-\frac{1}{2} \tan ^{-1}\left(\frac{\ln x}{x}\right)+C$

$\Rightarrow I=\frac{1}{4} \ln \left|\frac{\ln x-x}{\ln x+x}\right|-\frac{1}{2} \tan ^{-1}\left(\frac{\ln x}{x}\right)+C$