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.
 A: First we notice that$$\int_{0}^{1}x^{y}dy=\dfrac{x-1}{\ln x}.$$Therefore then we have$$\begin{aligned}
\int_{0}^{1}\dfrac{x^{2}-x}{(1+x)\ln x}dx&=\int_{0}^{1}\dfrac{x}{1+x}\cdot\dfrac{x-1}{\ln x}dx\\
&=\int_{0}^{1}\dfrac{x}{1+x}\left(\int_{0}^{1}x^{y}dy\right)dx\\
&=\int_{0}^{1}\int_{0}^{1}\dfrac{x^{y+1}}{1+x}dydx\\
&=\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}(-1)^{n}x^{n+y+1}dydx\\
&=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{1}\int_{0}^{1}x^{n+y+1}dxdy\\
&=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{1}\dfrac{1}{n+y+2}dy\\
&=\sum_{n=0}^{\infty}(-1)^{n}(\ln(n+3)-\ln(n+2))\\
&=\ln\big{(}\prod_{n=0}^{\infty}\dfrac{(2n+3)^{2}}{(2n+2)(2n+4)}\big{)}\\
&=\ln2+\ln\big{(}\prod_{n=0}^{\infty}\dfrac{(2n+1)(2n+3)}{(2n+2)^{2}}\big{)}\\
&=\ln2+\ln\dfrac{2}{\pi}=\ln\dfrac{4}{\pi}.\text{(Wallis product)}\\
\end{aligned}$$
A: 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:$$I=\ln\frac{3}{2}-\ln\frac{4}{3}+\ln\frac{5}{4}-\ln\frac{6}{5}+\cdots=\ln\frac{9}{8}+\ln\frac{25}{24}+\ln\frac{49}{48}+\cdots =\ln\frac{4}{\pi}.$$
A: To the nice posted solutions - just to add the third approach :)
We can evaluate a more general integral
$$\boxed{\,I(a)=\int_0^1\frac{x^a(x-1)}{(1+x)\ln x}dx\,;\,\,a>-1\,}$$
The integrand does not have singularities on the interval; taking the derivative with respect to $a$
$$I'(a)=\int_0^1\frac{x^a(x-1)}{(1+x)}dx=\int_0^1 x^adx-2\int_0^1\frac{x^a}{1+x}dx=\frac{1}{1+a}+J$$
where
$$J=-2\int_0^1(1-x+x^2-x^3+...)x^a\,dx=-2\Big(\frac{1}{1+a}-\frac{1}{2+a}+\frac{1}{3+a}-...\Big)$$
$$=-2\lim_{N\to\infty}\Big(\sum_{k=1}^N\frac{1}{k+a}-\sum_{k=1}^{N/2}\frac{1}{k+a/2}\Big)$$
Given that $\psi(1+a)=-\gamma-\lim_{N\to\infty}\sum_{k=1}^N \Big(\frac{1}{k+a}-\frac{1}{k}\Big)\,\,;\,\,\psi(x)$ - digamma-function.
$$J(a)=-2\Big(-\psi(1+a)+\psi(1+a/2)+\lim_{N\to\infty}\sum_{k=N/2+1}^N\frac{1}{k}\Big)=2\Big(\psi(1+a)-\psi(1+a/2)-\ln2\Big)$$
Using $\psi(x)=\frac{d}{dx}\ln\Gamma(x)$ and taking the antiderivative
$$I(a)=\int^aJ(x)dx+C=\ln(1+a)+2\ln\Gamma(1+a)-4\ln\Gamma(1+a/2)-(2\ln2)a+C$$
where $C$ is some constant. To define $C$ we see that at $a\to\infty\,\, I(a)\to0$. Using the asymptotics $\Gamma(1+a)=\sqrt{2\pi a}\Big(\frac{a}{e}\Big)^a\,$ at $\,a\to\infty$
$$I(a)\sim\ln a-2\ln\sqrt{2\pi}+\ln a-2a\ln a-2a$$
$$+\ln a+4\ln\sqrt{2\pi}-2\ln (a/2)-2a\ln (a/2)+2a-(2\ln2)a+C$$
$$I(a\to\infty)\to-2\ln\sqrt{2\pi}+2\ln2+C=\ln\frac{2}{\pi}+C=0\,\,\Rightarrow\,\,\boxed{\,\,C=\ln\frac{\pi}{2}\,\,}$$
Finally, we get for $I(a)$
$$\boxed{\boxed{\,\,I(a)=\ln(1+a)+2\ln\frac{\Gamma(1+a)}{\Gamma^2(1+a/2)}-(2\ln2)a+\ln\frac{\pi}{2}\,;\,\,a>-1\,\,}}$$
For several integer values of $a$ we get
$$I(1)=\ln 2-4\ln\frac{\sqrt{\pi}}{2}-2\ln2+\ln\frac{\pi}{2}\,\,\Rightarrow\,\,\boxed{\,\,I(1)=\ln\frac{4}{\pi}\,\,}$$
$$\boxed{\,\,I(0)=\ln\frac{\pi}{2}\,\,}$$
etc.
