Finding $\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{4-x^2}}dx$ 
Finding $$\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{4-x^2}}dx$$

Try: Let $$I=\frac{1}{2}\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{1-\frac{x^2}{4}}}=\frac{1}{2}\int^{1}_{0}\sum^{\infty}_{n=0}\binom{-1/2}{n}\bigg(-\frac{1}{4}\bigg)^nx^{2n}\ln^2(x)dx$$
$$I=\frac{1}{2}\sum^{\infty}_{n=0}\binom{-1/2}{n}\bigg(-\frac{1}{4}\bigg)^n\int^{1}_{0}x^{2n}\ln^2(x)dx$$
Using By parts , We have 
$$I=\frac{1}{2}\sum^{\infty}_{n=0}\binom{-1/2}{n}\bigg(-\frac{1}{4}\bigg)^n\frac{2}{(2n+1)^3}$$
But answer given as $\displaystyle\frac{7\pi^3}{216}$
I am not understand How can i get it. could some help me , Thanks 
 A: A sketch (this approach is generalizable to arbitrary even powers of the logarithm):
We use the standardbranch of $\log$ throughout
$$
\Re \log^2(1-e^{2 ix})=\log^2(2 \sin(x))-(x-\frac{\pi}{2})^2 \quad (\star)
$$
Integrating the LHS over a large rectangle in the upper half plane with verticies $\{(0,0),(\frac{\pi}{6},0),(\frac{\pi}{6},i\infty),(0,i\infty)\}$ we obtain (the integral over the imaginary line cancels, since we are only interested in real components)
$$
\Re\int_0^{\pi/6}\log^2(1-e^{2 ix})=-\Im\int_0^{\infty}\log^2(1-ae^{-2x})=-\Im(G(a))
$$
where $a=e^{-i \pi/3}$. Substituting $e^{-2x}=q$ we get
$$
2 G(a)=\int_0^1 \frac{\log^2(1-a q)}{q}
$$
using repeated integration by parts we get ($\text{Li}_n(z)$ denotes the Polylogarithm of order $n$)
$$
2 G(a)=-2\text{Li}_3(1-a)+2\text{Li}_2(1-a)\log(1-a)+\log(1-a)^2\log(a)+2\text{Li}_3(1)
$$
adding some Polylogarithm wizardy we find that (see Appendix)
$$
\Im(G(a))=-\color{blue}{\frac{\pi^3}{324}}
$$
using furthermore the trivial integral
$$
\int_0^{\pi/6}dx(x-\frac{\pi}{2})^2=\color{green}{\frac{19 \pi^3}{648}}
$$
we find indeed

$$
\int_0^{\pi/6}dx\log^2(2 \sin(x))=\\
\color{green}{\frac{19 \pi^3}{648}}+\color{blue}{\frac{\pi^3}{324}}=\frac{7\pi^3}{216}
$$

which is equivalent to the claim in question
Following OP we might rewrite the integral as an infinte sum, which gives us the hardly to believe corollary

$$
\sum_{n\geq 0}\frac{1}{16^n(2n+1)^3}\binom{2n}{n}=\frac{7\pi^3}{216}
$$

Appendix
The functional equations of the Dilogarithm  immediately delievers
$$\Re\text{Li}_2(1-a)= \frac{1}{2}(\text{Li}_2(1-a)+\text{Li}_2(1-a^{-1})\\=-\frac{1}{4}\log^2(a)=\frac{\pi^2}{36}$$
The Trilogarithm part is a bit trickier,
$\Im\text{Li}_3(1-z)=\Im\int_0^{1-z}\text{Li}_2(x)/x=-\Im\text{Li}_3(z)$ together with the inversion formula  
 $\text{Li}_3(-z)-\text{Li}_3(-1/z)=-\frac{1}{6}(\log^3(z)+\pi^2\log(z))$gives us that
$$
\Im\text{Li}_3(1-a)=\frac{5\pi^3}{162}
$$
since $\log(1-a)=\frac{i\pi}{3}$ we find
$$
2\Im(G(a))=-2\frac{5\pi^3}{162}+2\frac{\pi^2}{36}\frac{\pi}{3}+\frac{\pi^2}9\frac{\pi}3
$$
or 
$$
\Im(G(a))=-\frac{\pi^3}{324}
$$
A: We will integrate $\,f_p(x) := \ln^2(x)/\sqrt{p-x^2}\,$ using hypergeometric functions. Let $\,y:=x^2/p\,,\,$ and
 $\,h_0(x) := \!_1F_0(\frac12;;y) = 1/\sqrt{1-y},\,$
 $\,h_1(x) := \!_2F_1(\frac12,\frac12;\frac32;y),\,$
 $\,h_2(x) := \!_3F_2(\frac12,\frac12,\frac12;\frac32,\frac32;y),\,$ and
 $\,h_3(x) := \!_4F_3(\frac12,\frac12,\frac12,\frac12;\frac32,\frac32,\frac32;y).\,$ Note that $\,x h_n^{'}(x) = h_{n-1}(x)-h_n(x)\,$ if $\,n>0.\,$
Let $\,g_p(x) := \int f_p(x)\,dx = \sqrt{y}\,(2h_3(x) -2\ln(x)h_2(x) +\ln(x)^2h_1(x)).\,$ Now $\,g_p(0) = 0,\,$ and thus
$\, I:=\int_0^1 f_p(x)\,dx = g_p(1) = 2h_3(1/p)/\sqrt{p}.\,$
In our case $\,p=4\,$ and $\,I = h_3(1/4) = 7\pi^3/216.\,$
A: This is as far as I've gotten with the integral. I will admit, I underestimated the difficulty of this integral. The four in the denominator proved to be a lot more of a nuisance than I thought. Nevertheless, this integral can be painstakingly computed by hand with the help of our best friend

$$\log\sin x=-\sum\limits_{n\geq1}\frac {\cos 2nx}n-\log 2\tag1$$

First, we make the substitution $x\mapsto 2\sin x$ to clear the four in the denominator$$\begin{align*}I & =\int\limits_0^{\pi/6}dx\,\log^2(2\sin x)\\ & =\int\limits_0^{\pi/6}dx\,\log^22+\log 4\int\limits_0^{\pi/6}dx\,\log\sin x+\int\limits_0^{\pi/6}dx\,\log^2\sin x\tag2\end{align*}$$Call the remaining integrals in (2) $I_1$, $I_2$, and $I_3$ respectively. The first integral $I_1$ is trivial
$$I_1\color{blue}{=\frac {\pi}6\log^22}\tag3$$
The second and third integrals $I_2$ and $I_3$ don't have to be fully computed to simplify the result. First, let's tackle $I_2$. Using the expansion for (1), we get
$$\begin{align*}I_2 & =-\log 4\int\limits_0^{\pi/6}dx\,\sum\limits_{n\geq1}\frac {\cos 2nx}n+\log 2\\ & \color{red}{=-\frac {\pi}3\log^22-\log 4\int\limits_0^{\pi/6}dx\,\sum\limits_{n\geq1}\frac {\cos 2nx}n}\tag4\end{align*}$$
Leave $I_2$ as is, because we can expand $I_3$ courtesy of the square and find out that a portion of $I_3$ cancels out with $I_1+I_2$. Doing the math gives
$$\begin{align*}I_3 & \color{brown}{=\sum\limits_{n\geq1}\sum\limits_{m\geq1}\int\limits_0^{\pi/6}dx\,\frac {\cos 2mx\cos 2nx}{mn}+\log 4\int\limits_0^{\pi/6}dx\,\sum\limits_{n\geq1}\frac {\cos 2nx}n+\frac {\pi}6\log^22}\tag5\end{align*}$$
Immediately, notice how the two infinite series cancel out, leaving us with a much more nicer sum to deal with. Adding (3), (4), and (5) together and canceling out all the terms leaves us with
$$I=\sum\limits_{n\geq1}\sum\limits_{m\geq1}\int\limits_0^{\pi/6}dx\,\frac {\cos 2mx\cos 2nx}{mn}$$
Now, what's left is to show that the nested sum actually equals $\frac {7\pi^3}{216}$.
