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 

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 $\frac {7\pi^3}{216}$.
