If the value of the definite integral ${\int }_0^1 \frac{{\sin }^{-1}\sqrt{x}}{x^2-x+1}dx$ is $\frac{{\pi }^2}{\sqrt{n}}$ (where $n\in N)$), then the value of n is __.

$\begin{array}{l}I={\int }_0^1 \frac{{\sin }^{-1}\sqrt{x}}{x^2-x+1}dx\\ I={\int }_0^1 \frac{{\sin }^{-1}\sqrt{1-x}}{x^2-x+1}dx={\int }_0^1 \frac{{\cos }^{-1}\sqrt{x}}{x^2-x+1}dx\end{array}$

On adding (1) and (2), we get

$\begin{array}{l}2I={\int }_0^1 \frac{{\sin }^{-1}\sqrt{x}+{\cos }^{-1}\sqrt{x}}{x^2-x+1}dx\\ =\frac{\pi }{2}{\int }_0^1 \frac{dx}{x^2-x+1}dx\\ =\frac{\pi }{2}{\int }_0^1 \frac{dx}{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx\\ \therefore 2I=\frac{\pi }{2}\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}{\left[{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)\right]}_0^1=\frac{{\pi }^2}{3\sqrt{3}}\end{array}$

Hence, $I=\frac{{\pi }^2}{6\sqrt{3}}=\frac{{\pi }^2}{\sqrt{108}}\equiv \frac{{\pi }^2}{\sqrt{n}}$