The function $y=f(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}$ in $(-1,1)$ satisfying $f(0)=0$. Then $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}f(x)d(x)$ is

$\dfrac{dy}{dx} + \dfrac{x}{x^2-1}\,y = \dfrac{x^4+2x}{\sqrt{1-x^2}}, \quad x \in (-1,1)$ $f(0) = 0$ $\text{To find: } \int_{-\sqrt{3}/2}^{\sqrt{3}/2} f(x)\,dx$ $1.\ \text{integrating factor } \mu$ $2.\ f(0)=0 \Rightarrow f(x)$ $3.\ \int_{-\sqrt3/2}^{\sqrt3/2} f\,dx \text{ by symmetry + } x=\sin\theta$ $\mu = e^{\int \frac{x}{x^2-1}\,dx} = e^{\frac{1}{2}\ln|x^2-1|} = \sqrt{1-x^2} \quad (|x| < 1)$ $\dfrac{d}{dx}\!\left[y\sqrt{1-x^2}\right] = x^4+2x \Rightarrow y\sqrt{1-x^2} = \dfrac{x^5}{5} + x^2 + C$ $C = 0 \Rightarrow f(x) = \dfrac{\tfrac{1}{5}x^5 + x^2}{\sqrt{1-x^2}}$ $\int_{-\sqrt3/2}^{\sqrt3/2} f(x)\,dx = 2\int_0^{\sqrt3/2} \dfrac{x^2}{\sqrt{1-x^2}}\,dx \quad (\text{odd part} \to 0)$ $x = \sin\theta: \ \int_0^{\pi/3} \sin^2\theta\,d\theta = \left[\dfrac{\theta}{2} - \dfrac{\sin 2\theta}{4}\right]_0^{\pi/3} = \dfrac{\pi}{6} - \dfrac{\sqrt{3}}{8}$ $2\!\left(\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{8}\right) = \dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4} \Rightarrow \text{option } (2)$