$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$

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 $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx = \frac{\pi^2}{6\sqrt{3}}$$

$$ \begin{align} \int_0^1\frac{\sin^{-1}\left(\sqrt{x}\right)}{x^2-x+1}\,\mathrm{d}x &=\int_0^1\frac{\frac12\cos^{-1}(1-2x)}{x^2-x+1}\,\mathrm{d}x\tag{1}\\ &=\int_{-1}^1\frac{\cos^{-1}(x)}{3+x^2}\,\mathrm{d}x\tag{2}\\ &=\frac12\int_{-1}^1\frac{\cos^{-1}(x)+\cos^{-1}(-x)}{3+x^2}\,\mathrm{d}x\tag{3}\\ &=\frac\pi2\int_{-1}^1\frac1{3+x^2}\,\mathrm{d}x\tag{4}\\ &=\frac\pi{2\sqrt3}\int_{-1/\sqrt3}^{1/\sqrt3}\frac1{1+x^2}\,\mathrm{d}x\tag{5}\\ &=\frac\pi{\sqrt3}\tan^{-1}\left(\frac1{\sqrt3}\right)\tag{6}\\ &=\frac{\pi^2}{6\sqrt3}\tag{7} \end{align} $$ Explanation:
$(1)$: $\sin^{-1}\left(\sqrt{x}\right)=\frac12\cos^{-1}(1-2x)$
$(2)$: substitute $x\mapsto\frac{1-x}2$
$(3)$: $\int_{-1}^1f(x)\,\mathrm{d}x=\frac12\int_{-1}^1(f(x)+f(-x))\,\mathrm{d}x$
$(4)$: $\cos^{-1}(x)+\cos^{-1}(-x)=\pi$
$(5)$: substitute $x\mapsto\sqrt3x$
$(6)$: integrate
$(7)$: simplify

Let, $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt x)}{x^2-x+1}dx\tag 1$$ Now, using property of definite integral $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we get

$$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{(1-x)^2-(1-x)+1}dx$$ $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx\tag 2$$ Now, adding (1) & (2), we get $$I+I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})}{x^2-x+1}dx+\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$ $$2I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})+\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$

$$I=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x}\sqrt{1-1+x}+\sqrt{1-x}\sqrt{1-x})}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(x+(1-x)}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(1)}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\frac{\pi}{2}}{x^2-x+1}dx$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{x^2-x+1}$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}}$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt 3}{2}\right)^2}$$

$$=\frac{\pi}{4}\left[\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{\left(x-\frac{1}{2}\right)}{\frac{\sqrt 3}{2}}\right)\right]_{0}^{1}$$ $$=\frac{\pi}{2\sqrt 3}\left[\frac{\pi}{6}+\frac{\pi}{6}\right]$$ $$=\frac{\pi^2}{6\sqrt 3}$$