Evaluating $\int_{0}^{1}\frac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$ Find this integral
$$\operatorname I=\int\limits_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$$
My try: let
$$f(x)=x^4-2x^3+2x^2-x+1$$
I found 
$$f(1-x)=(1-x)^4-2(1-x)^3+2(1-x)^2-x+1=x^4-2x^3+2x^2-x+1=f(x)$$
so
$$I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$
so
$$2I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}+\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$
then I can't,Thank you very much
 A: Firstly, as shown in @Ron Gordon's answer, we should eliminate the $\arcsin(\sqrt{x})$ in the numerator. Take the substitution that $\arcsin(\sqrt{x})=t$, then
\begin{equation}
I=\int_0^{\pi/2}\frac{t\sin(2t)}{\sin(t)^8-2\sin(t)^6+2\sin(t)^4-\sin(t)^2+1}dt\\
=\int_0^{\pi/2}\frac{t\sin(2t)}{115/128+1/128\cos(8t)+3/32\cos(4t)}dt\\
\end{equation}
By taking $t'=\pi/2-t$, we can check
\begin{equation}
I=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin(2t)}{115/128+1/128\cos(8t)+3/32\cos(4t)}dt\\
=\frac{\pi}{4}\int_0^1\frac{1}{x^4-2x^3+2x^2-x+1}dx\\
=\frac{\pi}{4}\int_0^1\frac{1}{(x^2-x+\frac{1}{2})^2+\frac{3}{4}}dx\\
=\frac{\pi}{4}\int_0^1\frac{1}{((x-\frac{1}{2})^2+\frac{1}{4})^2+\frac{3}{4}}dx\\
\stackrel{y=x-1/2}{=}\frac{\pi}{4}\int_{-1/2}^{1/2}\frac{1}{(y^2+\frac{1}{4})^2+\frac{3}{4}}dy
\end{equation}
Also, I do not have some good manner to solve it but to factorize the denominator in complex. We can easy to get the 4 roots of the denominator that
\begin{equation}
x_{1,2}=\pm\frac{1}{2}\sqrt{-1-2i\sqrt{3}}\\
x_{3,4}=\pm\frac{1}{2}\sqrt{-1+2i\sqrt{3}}
\end{equation}
Then, you can factorize the last fraction by solve the following problem:
\begin{equation}
\frac{C_1x+C_2}{x-x_1}+\frac{C_3x+C_4}{x-x_2}+\frac{C_5x+C_6}{x-x_3}+\frac{C_7x+C_8}{x-x_4}=\frac{1}{(y^2+\frac{1}{4})^2+\frac{3}{4}}
\end{equation}
By doing this, you can get the solution of $I$. Since the result is too long I ignore them. The approximation of the result is $0.9095208091$. I think there must be some easier method to deal with the integral of rational fraction. 
