Let the solution curve y = f(x) of the differential equation...

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

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

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

So, solution of the differential equation is

$y \cdot \sqrt{1 - {x}^{2}} = ∫ \dfrac{{x}^{4} + 2x}{\sqrt{1 - {x}^{2}}} \cdot \sqrt{1 - {x}^{2}}dx$

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

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

Since the curve passes through origin

so $x = 0, y = 0$ $⇒ C = 0$

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

Now,

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

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

Let $I = ∫ \dfrac{{x}^{2}}{\sqrt{1 - {x}^{2}}}dx$

Assume $1 - {x}^{2} = {t}^{2} ⇒ 1 - {t}^{2} = {x}^{2}$

and $- 2tdt = 2xdx ⇒ dx = - \dfrac{tdt}{\sqrt{1 - {t}^{2}}}$

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

$= - ∫ \sqrt{1 - {t}^{2}}dt$

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

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

$⇒ {∫}_{\dfrac{- \sqrt{3}}{2}}^{\dfrac{\sqrt{3}}{2}}f\left(x\right)dx$

$= 2{\left[- \dfrac{x}{2}\sqrt{1 - {x}^{2}} - \dfrac{1}{2}\sin^{- 1}\sqrt{1 - {x}^{2}}\right]}_{0}^{\dfrac{\sqrt{3}}{2}}$

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

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