Let y = f(x) be the solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{xy}}{{\mathrm{x}}^2-1}=\frac{{\mathrm{x}}^6+4\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}},-1<\mathrm{x}<1$ such that f(0)=0. If $6{\int }_{-1/2}^{1/2} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=2\mathrm{\pi }-\mathrm{\alpha }$ then α² is equal to ________.

$\begin{array}{r}\text{ I.F. }{\mathrm{e}}^{-\frac{1}{2}\int \frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}}={\mathrm{e}}^{-\frac{1}{2}\ln \left(1-{\mathrm{x}}^2\right)}=\sqrt{1-{\mathrm{x}}^2}\\ \mathrm{y}\times \sqrt{1-{\mathrm{x}}^2}=\int \left({\mathrm{x}}^6+4\mathrm{x}\right)\mathrm{dx}=\frac{{\mathrm{x}}^7}{7}+2{\mathrm{x}}^2+\mathrm{c}\end{array}$
Given y(0) = 0 $\Rightarrow$ c = 0
$\mathrm{y}=\frac{\frac{{\mathrm{x}}^7}{7}+2{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}$
Now, $6{\int }_{-\frac{1}{2}}^{\frac{1}{2}} \frac{\frac{{\mathrm{x}}^7}{7}+2{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}=6{\int }_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$
$=24{\int }_0^{\frac{1}{2}} \frac{{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}$
Put $\mathrm{x}=\sin \mathrm{\theta }$
$\mathrm{dx}=\cos \mathrm{\theta d\theta }$
$\begin{array}{l}=24{\int }_0^{\frac{\mathrm{\pi }}{6}} \frac{{\sin }^2\mathrm{\theta }}{\cos \mathrm{\theta }}\cos \mathrm{\theta d\theta }\\ =24{\int }_0^{\frac{\mathrm{\pi }}{6}} \left(\frac{1-\cos 2\mathrm{\theta }}{2}\right)\mathrm{d\theta }=12{\left[\mathrm{\theta }-\frac{\sin 2\mathrm{\theta }}{2}\right]}_0^{\frac{\mathrm{\pi }}{6}}\\ =12\left(\frac{\mathrm{\pi }}{6}-\frac{\sqrt{3}}{4}\right)\\ =2\mathrm{\pi }-3\sqrt{3}\\ {\mathrm{\alpha }}^2=(3\sqrt{3}{)}^2=27\end{array}$