\( \int \frac{x^2\left ( \ln x-1 \right )}{x^4-\ln^4 x}\,dx \)

[Tolaso J Kos]

Evaluate the integral: $$\int \frac{x^2\left ( \ln x-1 \right )}{x^4-\ln^4 x}\,dx$$

[Papapetros Vaggelis]

For each \(\displaystyle{x>0}\) holds:

\(\displaystyle{x^4-\ln^4\,x=\left(x^2-\ln^2\,x\right)\,\left(x^2+\ln^2\,x\right)=\left(x-\ln\,x\right)\,\left(x+\ln\,x\right)\,\left(x^2+\ln^2\,x\right)}\)
with \(\displaystyle{x^2+\ln^2\,x>0}\) , \(\displaystyle{x-\ln\,x>0}\) cause \(\displaystyle{\ln\,x\leq x-1<x}\) .

Also, the function \(\displaystyle{f:\left(0.+\infty\right)\longrightarrow \mathbb{R}\,,f(x)=\ln\,x+x}\) is continuous and strictly increasing on \(\displaystyle{\left(0,+\infty\right)}\) with

\(\displaystyle{f\,\left(\left(0,+\infty\right)\right)=\left(\lim_{x\to 0^{+}}f(x),\lim_{x\to +\infty}f(x)\right)=\left(-\infty,+\infty\right)}\) .

So, the function \(\displaystyle{f}\) has a unique root \(\displaystyle{x_0>0}\) .

Now, we integrate either on \(\displaystyle{\left(0,x_0\right)}\) or on \(\displaystyle{\left(x_0,+\infty\right)}\).

\(\displaystyle{\begin{aligned}I&=\int \dfrac{x^2\,\left(\ln\,x-1 \right )}{x^4-\ln^4\,x}\,\mathrm{d}x\\&=\int \dfrac{\displaystyle{\dfrac{\ln\,x-1}{x^2}}}{1-\displaystyle{\left(\dfrac{\ln\,x}{x} \right )^4}}\,\mathrm{d}x\\&=\int \dfrac{1}{\displaystyle{\left(\dfrac{\ln\,x}{x} \right )^4-1}}\,\mathrm{d}\,\left(\dfrac{\ln\,x}{x} \right )\\&=\int \dfrac{1}{\displaystyle{\left(\dfrac{\ln\,x}{x}-1 \right )\left(\dfrac{\ln\,x}{x}+1 \right )\left(\left(\dfrac{\ln\,x}{x} \right )^2+1 \right )}}\,\mathrm{d}\,\left(\dfrac{\ln\,x}{x} \right )\end{aligned}}\)

By substituting \(\displaystyle{u=\dfrac{\ln\,x}{x}}\), we have :

\(\displaystyle{I=\int \dfrac{1}{\left(u-1\right)\,\left(u+1\right)\,\left(u^2+1\right)}\,\mathrm{d}u}\)

and the calculation procedure of the last integral is quite known ( partial decomposition) .

Finally we get :

\(\displaystyle{I=\dfrac{1}{4}\,\left[\ln\,\left(x-\ln\,x\right)-\ln\,\left|x+\ln\,x\right|-2\,\arctan\,\dfrac{\ln\,x}{x}\right]+c\,,c\in\mathbb{R}}\) .