Show that $\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx=\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx=\pi$ Two integrals exhibit the same closed form
Motivated by this interesting question.

$$\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right)\mathrm dx=\pi\tag1$$

and 

$$\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right)\mathrm dx=\pi\tag2$$

Making an attempt
$$u={x\over 1-x}\implies -(1-x)^2du=dx\tag a$$
$$u=\sqrt{x\over 1-x}\implies 2x(1-x)^2du=dx\tag b$$
I try on $(a)$ and $(b)$ it doesn't seem to be working. So I try another
$u=\sqrt{1-x}\implies -2\sqrt{1-x}du=dx$, apply to $(1)$ then we have 
$$2\int_{0}^{1}\sqrt{1-u^2}\ln\left({1-u^2\over u^2}\right)\mathrm du=I_1-I_2\tag3$$
$$I_1-I_2=2\int_{0}^{1}\sqrt{1-u^2}\ln(1-u^2)\mathrm du-2\int_{0}^{1}\sqrt{1-u^2}\ln u^2\mathrm du\tag4$$
For $I_1$ make $v=\sqrt{1-u^2}\implies -{\sqrt{1-u^2}\over u}dv=du$, then 
$$I_1=4\int_{0}^{1}{v^2\over \sqrt{1-v^2}}\cdot\ln v \mathrm dv\tag5$$
Recall $${1\over \sqrt{1+x}}=\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}(-x)^k\tag6$$
then $(5)$ becomes
$$I_1=4\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\int_{0}^{1}v^4\ln v \mathrm dv\tag7$$
$$I_1=-\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\tag8$$
Apply same substitution, $I_2$ becomes
$$I_2=2\int_{0}^{1}{v^2\over \sqrt{1-v^2}}\cdot\ln(1-v^2)\mathrm dv\tag9$$
$$I_2=2\sum_{0}^{\infty}{(2k-1)!!\over (2k)!!}\int_{0}^{1}v^4\ln(1-v^2)\mathrm dv\tag c$$
Surely this is going wrong because there is not sign of $\pi $ anywhere in $I_1$ nor $I_2$!
How do we show that $(1)=(2)$ and prove one of them?
 A: The integrals are equal
Note that
$$
\int_0^1\sqrt\frac{x}{1-x}\ln\frac{x}{1-x}\,dx-\int_0^1x\sqrt\frac{x}{1-x}\ln\frac{x}{1-x}\,dx
=\int_0^1\sqrt{x(1-x)}\ln\frac{x}{1-x}\,dx.
$$
The last integral becomes zero, since $f(x)=\sqrt{x(1-x)}\ln\frac{x}{1-x}$ satisfies $f(x)+f(1-x)=0$.
The value of the integrals
To calculate the integrals, introduce
$$
I(a)=\int_0^1x^a(1-x)^{-a}\,dx.
$$
The integral in (1) is then just $I'(1/2)$. Now, however, $I(a)$ is a beta-integral that via the symmetry of the Gamma function reduces to
$$
I(a)=B(1+a,1-a)=\frac{\Gamma(1+a)\Gamma(1-a)}{\Gamma(2)}=a\Gamma(a)\Gamma(1-a)=\frac{a\pi}{\sin(a\pi)},
$$
and differentiating yields
$$
I'(1/2)=\frac{\pi}{\sin(\pi/2)}-\frac{\pi^2/2}{\sin^2(\pi/2)}{\cos(\pi/2)}=\pi.
$$
A: On the path of Latte,
$\displaystyle J=\int_0^1 \sqrt{\dfrac{x}{1-x}}\ln\left(\dfrac{x}{1-x}\right)dx$
Perform the change of variable $y=\sqrt{\dfrac{x}{1-x}}$ and then, use integration by parts,
$\begin{align} J&=4\int_0^{\infty} \dfrac{x^2\ln x}{(1+x^2)^2}dx\\
&=\left[-\dfrac{2x\ln x}{1+x^2}\right]_0^{\infty}+2\int_0^{\infty}\dfrac{\ln x}{1+x^2}dx+2\int_0^{\infty}\dfrac{1}{1+x^2}dx\\
&=2\int_0^{\infty}\dfrac{1}{1+x^2}dx\\
&=2\Big[\arctan x\Big]_0^{\infty}\\
&=2\times \dfrac{\pi}{2}\\
&=\boxed{\pi}
\end{align}$
$\begin{align}
\int_0^{\infty}\dfrac{\ln x}{1+x^2}dx=\int_0^{1}\dfrac{\ln x}{1+x^2}dx+\int_1^{\infty}\dfrac{\ln x}{1+x^2}dx
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}
\int_0^{\infty}\dfrac{\ln x}{1+x^2}dx&=\int_0^{1}\dfrac{\ln x}{1+x^2}dx-\int_0^{1}\dfrac{\ln x}{1+x^2}dx\\
&=0
\end{align}$
A: Let
$$I=\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx,J=\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx. $$
Then
$$ I-J= \int_{0}^{1}\sqrt{x(1-x)}\cdot\ln\left({x\over 1-x}\right) dx.$$
Using $x\to1-x$, it is easy to get
$$ I-J=-(J-I) $$
and hence $I-J=0$ or $I=J$. The rest follows @mickep's answer.
