Mathematics Q6

Similar method with Akhil Jain 's

$\begin{aligned} I & = \int_0^1 \frac {\arcsin \sqrt x}{x^2-x+1} dx & \small \color{#3D99F6} \text{Let } \sin \theta = \sqrt x, \ \sin^2 \theta = x, \ dx = 2\sin \theta \cos \theta d\theta \\ & = \int_0^\frac \pi 2 \frac {2\theta \sin \theta \cos \theta}{\sin^4 \theta - \sin^2 \theta + 1} d\theta & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac {2\theta \sin \theta \cos \theta}{\sin^4 \theta - \sin^2 \theta + 1} + \frac {2 \left(\frac \pi 2 - \theta \right) \cos \theta \sin \theta}{\cos^4 \theta - \cos^2 \theta + 1} d\theta \\ & = \frac 12 \int_0^\frac \pi 2 \frac {2\theta \sin \theta \cos \theta}{\sin^4 \theta - \sin^2 \theta + 1} + \frac {\left(\pi - 2 \theta \right) \sin \theta \cos \theta}{\sin^4 \theta - \sin^2 \theta + 1} d\theta \\ & = \frac \pi 2 \int_0^\frac \pi 2 \frac {\sin \theta \cos \theta}{\sin^4 \theta - \sin^2 \theta + 1} d\theta & \small \color{#3D99F6} \text{Let }u = \sin \theta, \ du = \cos \theta \ d\theta \\ & = \frac \pi 2 \int_0^1 \frac u{u^4 - u^2 + 1} du & \small \color{#3D99F6} \text{Let }t = u^2, dt = 2u \ du \\ & = \frac \pi 4 \int_0^1 \frac 1{t^2 - t + 1} dt \\ & = \frac \pi 4 \int_0^1 \frac 1{\left(t - \frac 12\right)^2 + \frac 34} dt \\ & = \frac \pi 3 \int_0^1 \frac 1{\frac 43 \left(t - \frac 12\right)^2 + 1} dt & \small \color{#3D99F6} \text{Let } v = \frac 2{\sqrt 3} \left(t - \frac 12 \right), \ dv = \frac 2{\sqrt 3} dt \\ & = \frac \pi {2\sqrt 3} \int_{-\frac 1{\sqrt 3}}^{\frac 1{\sqrt 3}} \frac 1{v^2 + 1} dv \\ & = \frac \pi{2 \sqrt 3} \tan^{-1} v \ \bigg|_{-\frac 1{\sqrt 3}}^{\frac 1{\sqrt 3}} = \frac {\pi^2}{6 \sqrt 3} = \frac {\pi^2}{\sqrt{108}} \end{aligned}$

$\implies n = \boxed{108}$

Rather than substitution directly apply a+b-x property and use the formula of arcsina+arcsinb to get the integral of 1/x^2-x+1

$I=\int_0^1 \dfrac{\arcsin(\sqrt{x})}{1-x(1-x)}dx\\ I= \int_0^1\dfrac{\arcsin(\sqrt{1-x})}{1-x(1-x)}dx\\ 2I=\int_0^1 \dfrac{\arcsin(\sqrt{x})+\arcsin(\sqrt{1-x})}{1-x(1-x)}dx\\ \arcsin(\sqrt{x})+\arcsin(\sqrt{1-x})=\arcsin(\sin(\arcsin(\sqrt{x})+\arcsin(\sqrt{1-x})))\\ =\arcsin(\sin(\arcsin(\sqrt{x}))\cos(\arcsin(\sqrt{1-x}))+\sin(\arcsin(\sqrt{1-x}))\cos(\arcsin(\sqrt{x})))\\ =\arcsin(x+1-x)=\frac{\pi}{2}\\ I= \dfrac{\pi}{4} \int_0^1 \dfrac{1}{1-x(1-x)}dx= \dfrac{\pi}{4} \int_0^1 \dfrac{1}{(x-1/2)^2+3/4}dx\\ = \dfrac{\pi}{4} \dfrac{2}{\sqrt{3}}\arctan\left(\dfrac{2(x-1)}{\sqrt{3}}\right) |_0^1=\dfrac{\pi^2}{6\sqrt{3}}\\ =\dfrac{\pi^2}{\sqrt{\boxed{108}}}$