Let the solution curve $y=f(x)$ of the differential equation $\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}$, $x\in(-1,1)$ pass through the origin. Then $\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$ is equal to

<p>$${{dy} \over {dx}} + {{xy} \over {{x^2} - 1}} = {{{x^4} + 2x} \over {\sqrt {1 - {x^2}} }}$$</p> <p>which is first order linear differential equation.</p> <p>Integrating factor $(I.F.) = {e^{\int {{x \over {{x^2} - 1}}dx} }}$</p> <p>$= {e^{{1 \over 2}\ln |{x^2} - 1|}} = \sqrt {|{x^2} - 1|}$</p> <p>$= \sqrt {1 - {x^2}}$</p> <p>$\because$ $x \in ( - 1,1)$</p> <p>Solution of differential equation</p> <p>$y\sqrt {1 - {x^2}} = \int {({x^4} + 2x)dx = {{{x^5}} \over 5} + {x^2} + c}$</p> <p>Curve is passing through origin, $c = 0$</p> <p>$y = {{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}$</p> <p>$$\int\limits_{{{ - \sqrt 3 } \over 2}}^{{{\sqrt 3 } \over 2}} {{{{x^5} + 5{x^2}} \over {5\sqrt {1 - {x^2}} }}dx = 0 + 2\int\limits_0^{{{\sqrt 3 } \over 2}} {{{{x^2}} \over {\sqrt {1 - {x^2}} }}dx} } $$</p> <p>put $x = \sin \theta$</p> <p>$dx = \cos \theta \,d\theta$</p> <p>$$I = 2\int\limits_0^{{\pi \over 3}} {{{{{\sin }^2}\theta \,.\,\cos \theta d\theta } \over {\cos \theta }}} $$</p> <p>$= \int\limits_0^{{\pi \over 3}} {(1 - \cos 2\theta )d\theta }$</p> <p>$$ = \left. {\left( {\theta - {{\sin 2\theta } \over 2}} \right)} \right|_0^{{\pi \over 3}}$$</p> <p>$= {\pi \over 3} - {{\sqrt 3 } \over 4}$</p>