Let the solution curve y = f x of the differential equation d y d x + x y x 2 - 1 = x 4 + 2 x 1 - x 2 ,…

$\frac{dy}{dx}+\frac{xy}{x^2-1}=\frac{x^4+2x}{\sqrt{1-x^2}}$ is a linear differential equation.

Here $I.F.=e^{\int \frac{x}{x^2-1}dx}=e^{\frac{1}{2}\int \frac{2x}{x^2-1}dx}=e^{\frac{1}{2}\ln \left(x^2-1\right)}$

i.e. $I.F.=\sqrt{1-x^2}$$\quad \because x \in(-1,1)$

So, solution of the differential equation is 

$y·\sqrt{1-x^2}=\int \frac{x^4+2x}{\sqrt{1-x^2}}·\sqrt{1-x^2}dx$

$y\sqrt{1-x^2}=\int \left(x^4+2x\right)dx$

$y\sqrt{1-x^2}=\frac{x^5}{5}+x^2+C$

Since the curve passes through origin

so $x=0, y=0$ $\Rightarrow C=0$

i.e. $y=\frac{x^5}{5\sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}$

Now,

${\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx={\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{x^5}{5\sqrt{1-x^2}}dx+{\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{x^2}{\sqrt{1-x^2}}dx$

${\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx=0+2{\int }_0^{\frac{\sqrt{3}}{2}}\frac{x^2}{\sqrt{1-x^2}}dx$ (as $\frac{x^5}{5\sqrt{1-x^2}}$ is an odd function and$\frac{x^2}{\sqrt{1-x^2}}$  is an even function)

Let $I=\int \frac{x^2}{\sqrt{1-x^2}}dx$

Assume $1-x^2=t^2\Rightarrow 1-t^2=x^2$

and $-2tdt=2xdx\Rightarrow dx=-\frac{tdt}{\sqrt{1-t^2}}$

i.e. $I=\int \frac{1-t^2}{t}\left(-\frac{t}{\sqrt{1-t^2}}\right)dt$

$=-\int \sqrt{1-t^2}dt$

$=-\left[\frac{t}{2}\sqrt{1-t^2}+\frac{1}{2}{\sin }^{-1}t\right]+C$

$=-\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\sqrt{1-x^2}\right]+C$

$\Rightarrow {\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f\left(x\right)dx$

$=2{\left[-\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}{\sin }^{-1}\sqrt{1-x^2}\right]}_0^{\frac{\sqrt{3}}{2}}$

$=2\left(-\frac{\sqrt{3}}{8}-\frac{1}{2}{\sin }^{-1}\frac{1}{2}+\frac{1}{2}{\sin }^{-1}1\right)$

$=-\frac{\sqrt{3}}{4}-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{2}=\frac{\pi }{3}-\frac{\sqrt{3}}{4}$