$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$
$\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?

Let $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx.............(1)$$
Now Replace $x\rightarrow (1-x)\;,$ We get
$$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{(1-x)^2-(1-x)+1}dx = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{x^2-x+1}dx$$
So we get $$\displaystyle I = \int_{0}^{1}\frac{\cos^{-1}\sqrt{x}}{x^2-x+1}dx...............(2)$$
Above we have use the formula $$\bullet\; \displaystyle \sin^{-1}(\sqrt{x})+\sin^{-1}\sqrt{1-x} = \sin^{-1}(\sqrt{x})+\cos^{-1}(\sqrt{x}) = \frac{\pi}{2}$$
Now Add these two equations, We get
$$\displaystyle 2I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}+\cos^{-1}\sqrt{x}}{x^2-x+1} = \frac{\pi}{2}\int_{0}^{1}\frac{1}{x^2-x+1}dx = \frac{\pi}{2}\int_{0}^{1}\frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx$$
Now Put $\displaystyle \left(x-\frac{1}{2}\right) = t$ and $dx = dt$ and Changing Limt, We get
$$\displaystyle 2I = \int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{t^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = 2\cdot \frac{\pi}{2}\int_{0}^{\frac{1}{2}} \frac{1}{t^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = \frac{2\pi}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi^2}{3\sqrt{3}}$$
So we get $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx = \frac{\pi^2}{6\sqrt{3}}$$
