Evaluate $I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$ I am trying to calculate this integral:$$I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$$.
I tried to find the antiderivative but it didn't exist.
So i changed variable by set $t=\frac{x-1}{x+1}$ due to factor numerator is $x(x-1)$ and it led to:$$I=2\int_{-1}^{0}\frac{t^2+t}{(1-t)^3\ln{\frac{t+1}{1-t}}}dx$$ and it seems more harder.
The result from Wolfram Alpha is ok, but i don't know how to evaluate this result. Need some hints or advices from everyone. Thank you.

Another approach can be as below:
First, by this link https://en.wikipedia.org/wiki/Frullani_integral one can easily obtain: $$\int_0^1 \frac{x^{a-1}-x^{b-1}}{\ln x}dx=\int_0^\infty \frac{e^{-bt}-e^{-at}}{t}dt=\ln\frac{a}{b},$$
where $a,b\gt0.$
Second, we have a well-known relation that states: $$\frac{\pi}{2}=(\frac{2}{1}\times\frac{2}{3})(\frac{4}{3}\times\frac{4}{5})(\frac{6}{5}\times\frac{6}{7})(\frac{8}{7}\times\frac{8}{9})\times\cdots .   $$
For example you can see this relation in this link https://en.wikipedia.org/wiki/Pi .
Third, notice that: $\frac{1}{1+x}=1-x+x^2-x^3+\cdots  . $
Now, we get: $$I=\int_0^1\frac{(x^2-x)(1-x+x^2-x^3+\cdots)}{\ln x}dx=\sum_{k=0}^\infty (-1)^k\ln\frac{k+3}{k+2}.$$
The last sum is indeed:$$I=\ln\frac{3}{2}-\ln\frac{4}{3}+\ln\frac{5}{4}-\ln\frac{6}{5}+\cdots=\ln\frac{9}{8}+\ln\frac{25}{24}+\ln\frac{49}{48}+\cdots =\ln\frac{4}{\pi}.$$
