Computing $\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$

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.

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 $\sin^{-1}\sqrt{1-x}=\cos^{-1}\sqrt x$.

$$I=\pi\int_{0}^{\pi/2} \frac{\sin u}{4-\sin^2 u}= \pi \int_{0}^{\pi/2} \frac{\sin u}{3+\cos^2 u} du= \pi\int_{0}^{1} \frac{dv}{3+v^2}dv=\frac{\pi}{\sqrt{3}} \tan^{-1}(v/\sqrt{3})|_{0}^{1}= \frac{\pi^2}{6 \sqrt{3}}.$$ Lastly, we have used $v=\cos u$.