Computing $\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$ 
Compute:
$$\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$$
Answer: $\frac{\pi^2}{6\sqrt 3}$

My Attempt:
The obvious substitution: $\arcsin \sqrt x=t$. This transforms my integral (say $I$) to:
$$I=\int_0^{\frac{\pi}{2}} \frac{t\cdot 2\sin t\cos t}{\sin^4-\sin^2+1}dt$$
Then the substitution $t\rightarrow \left(\frac{\pi}{2}+0\right)-t$. This leads to the denominator remaining the same.
Then, I added the "versions" of $I$ and on some simplifications, obtained:
$$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\underbrace{\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}}_{f(t)}dt$$
Since $f(x)=f\left(\frac{\pi}{2}-x\right)$ we say:
$$I=\frac{\pi}{2}\int_0^{\frac{\pi}{4}}\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}dt$$
Now the substitution $2t=u$ transforms it to:
$$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\frac{\sin u}{1-\frac{\sin^2 u}{4}}du$$
Or,
$$I=\pi\int_0^{\frac{\pi}{2}}\frac{\sin u}{4-\sin^2 u}du$$
Then I tried to decompose it into partial fractions and integrate them individually, by converting them in terms of $\tan \frac u2$ but the answer I'm getting doesn't match the given one.
Thanks in advance!
 A: We could complete this integral with one integration-by-parts:
$$\int_0^1 \frac{\arcsin\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx = \frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\arcsin\sqrt{x}\Biggr|_0^1 - \frac{1}{\sqrt{3}}\int_0^1\frac{\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{x}\sqrt{1-x}}dx$$
$$= \frac{\pi^2}{6\sqrt{3}}-\frac{1}{\sqrt{3}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\arctan\left(\frac{2x}{\sqrt{3}}\right)}{\sqrt{\frac{1}{2}+x}\sqrt{\frac{1}{2}-x}}dx = \frac{\pi^2}{6\sqrt{3}}$$
where the second integral vanishes by odd symmetry.
A: Hint: Not all the substitutions work always. Also, you should first use $t=0+1-x$ then go for partial fraction $$I=\int_0^1 \frac{\sin^{-1} \sqrt x}{x^2-x+1} dx=\int_0^1 \frac{\sin^{-1} \sqrt {1-x}}{x^2-x+1} $$ Note that $\sin^{-1}\sqrt{1-x}=\cos^{-1}\sqrt x$.
A: $$I=\pi\int_{0}^{\pi/2} \frac{\sin u}{4-\sin^2 u}= \pi \int_{0}^{\pi/2} \frac{\sin u}{3+\cos^2 u} du= \pi\int_{0}^{1} \frac{dv}{3+v^2}dv=\frac{\pi}{\sqrt{3}} \tan^{-1}(v/\sqrt{3})|_{0}^{1}= \frac{\pi^2}{6 \sqrt{3}}.$$
Lastly, we have used $v=\cos u$.
