$${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}$$

which is first order linear differential equation.

Integrating factor $$(I.F.) = {e^{\int {{x \over {{x^2} - 1}}dx} }}$$

$$ = {e^{{1 \over 2}\ln |{x^2} - 1|}} = \sqrt {|{x^2} - 1|} $$

$$ = \sqrt {1 - {x^2}} $$

$$\because$$ $$x \in ( - 1,1)$$

Solution of differential equation

$$y\sqrt {1 - {x^2}} = \int {({x^4} + 2x)dx = {{{x^5}} \over 5} + {x^2} + c} $$

Curve is passing through origin, $$c = 0$$

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

$$\int\limits_{{{ - \sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {{{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}dx = 0 + 2\int\limits_0^{{{\sqrt 3 } \over 2}} {{{{x^2}} \over {\sqrt {1 - {x^2}} }}dx} } $$

put $$x = \sin \theta $$

$$dx = \cos \theta \,d\theta $$

$$I = 2\int\limits_0^{{\pi \over 3}} {{{{{\sin }^2}\theta \,.\,\cos \theta d\theta } \over {\cos \theta }}} $$

$$ = \int\limits_0^{{\pi \over 3}} {(1 - \cos 2\theta )d\theta } $$

$$ = \left. {\left( {\theta - {{\sin 2\theta } \over 2}} \right)} \right|_0^{{\pi \over 3}}$$

$$ = {\pi \over 3} - {{\sqrt 3 } \over 4}$$