The definite Integral involving natural logarithm Is there the solution for this integral
$$\int_{0}^{1}\frac{\left(\ln\left(\frac{1-t}{t}\right)\right)^2}{\sqrt{t-t^2}}dt$$
I tried to solve this integral by  Maclaurin series but the result was so coomplicated. Can you please help me. Thanks
 A: We have $$I=\int_{0}^{1}\frac{\log^{2}\left(\frac{1-t}{t}\right)}{\sqrt{t-t^{2}}}dt=\int_{0}^{1}\frac{\log^{2}\left(1-t\right)}{\sqrt{t-t^{2}}}dt-2\int_{0}^{1}\frac{\log\left(1-t\right)\log\left(t\right)}{\sqrt{t-t^{2}}}dt+\int_{0}^{1}\frac{\log^{2}\left(t\right)}{\sqrt{t-t^{2}}}dt
 $$ Now recalling the definition of the Beta function $$B\left(a,b\right)=\int_{0}^{1}t^{a-1}\left(1-t\right)^{b-1}dt
 $$ we have $$\int_{0}^{1}\frac{\log^{2}\left(1-t\right)}{\sqrt{t-t^{2}}}dt=\frac{\partial^{2}}{\partial b^{2}}B\left(a,b\right)_{a=1/2,\, b=1/2}\tag{1} $$ $$-2\int_{0}^{1}\frac{\log\left(1-t\right)\log\left(t\right)}{\sqrt{t-t^{2}}}dt=-2\frac{\partial^{2}}{\partial a \partial b}B\left(a,b\right)_{a=1/2,\, b=1/2}\tag{2}
 $$ and $$\int_{0}^{1}\frac{\log^{2}\left(t\right)}{\sqrt{t-t^{2}}}dt=\frac{\partial^{2}}{\partial a^{2}}B\left(a,b\right)_{a=1/2,\, b=1/2}\tag{3}. $$ Now it is sufficient to recall that $$B\left(a,b\right)=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}$$ and $$\Gamma'\left(s\right)=\Gamma\left(s\right)\psi\left(s\right)$$ where $\psi(s)$ is the digamma function.
Note that $(1)$ and $(3)$ are equal by simmetry so, since
$$\frac{\partial^{2}}{\partial b^{2}}B\left(a,b\right)_{a=1/2,\, b=1/2}=\frac{\partial^{2}}{\partial a^{2}}B\left(a,b\right)_{a=1/2,\, b=1/2}=\frac{\pi}{3}\left(3\gamma^{2}+\pi^{2}+6\gamma\psi\left(\frac{1}{2}\right)+3\psi\left(\frac{1}{2}\right)^{2}\right)
 $$ $$\frac{\partial^{2}}{\partial a\partial b}B\left(a,b\right)_{a=1/2,\, b=1/2}=\gamma^{2}\pi-\frac{\pi^{3}}{6}+2\gamma\pi\psi\left(\frac{1}{2}\right)+\pi\psi\left(\frac{1}{2}\right)^{2}
 $$ we have $$I=\color{red}{\pi^{3}}.$$
A: By using the substitutions $t=\frac{1}{1+u}$ and $u=v^2$we have:
$$ I=\int_{0}^{1}\log^2\left(\frac{1-t}{t}\right)\frac{dt}{\sqrt{t-t^2}}=\int_{0}^{+\infty}\frac{\log^2(u)}{(1+u)\sqrt{u}}\,du =8\int_{0}^{+\infty}\frac{\log^2(v)}{1+v^2}\,dv$$
that also is:
$$ I = 16\int_{0}^{1}\log^2(v)\sum_{k\geq 0}(-1)^k v^{2k}\,dk = 32\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^3}.$$
The last series is a famous series.
A: Set $t=\sin^2\theta$ . we have $dt=\sin2\theta\,d\theta$ and $\sqrt{t-t^2}=\frac{1}{2}\sin 2\theta$ as a result
$$I=\int_{0}^{1}\frac{\left(\ln\left(\frac{1-t}{t}\right)\right)^2}{\sqrt{t-t^2}}dt=\int_{0}^{1}\frac{\left(-\ln\left(\frac{t}{1-t}\right)\right)^2}{\sqrt{t-t^2}}dt=8\int_{0}^{\frac{\pi}{ 2}}\ln^2 \tan (\theta) d\theta$$
Set $x=\tan\theta$, we have
$$I=8\int_{0}^{\infty}\frac{\ln^2(x)}{1+x^2}dx=16\int_{0}^{1}\frac{\ln^2(x)}{1+x^2}dx$$
Set $x=e^{-y}$, we have
$$I=16\int_{0}^{\infty}\frac{y^2e^{-y}}{1+e^{-2y}}dy$$
Now you can use the  Maclaurin series .
