Let y=f(x) be the solution of the differential equation (d y)/(d x)+(x y)/(x^2-1)=(x^6+4 x)/(√(1-x^2)),-1<x<1 such that f(0)=0. If 6 ∫_-1 / 2^1 / 2…

Let $\displaystyle y=f(x)$ be the solution of the differential equation $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{x y}{x^2-1}=\frac{x^6+4 x}{\sqrt{1-x^2}},-1<x<1$ such that $\displaystyle f(0)=0$. If $\displaystyle 6 \int_{-1 / 2}^{1 / 2} f(x) \mathrm{d} x=2 \pi-\alpha$ then $\displaystyle \alpha^2$ is equal to $\displaystyle \_\_\_\_$.