The function $y=f(x)$ is the solution of the differential equation $\frac{d y}{d x}+\frac{x}{x^2-1} y=\frac{x^4+2 x}{\sqrt{1-x^2}}$ in $(-1,1)$ satisfying $f(0)=0$. Then, $\int\limits_{-\sqrt{3} / 2}^{\sqrt{3} / 2} f(x) d x$ is equal to

$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$

We have,

$\frac{d y}{d x}+\frac{x y}{x^2-1}=\frac{x^4+2 x}{\sqrt{1-x^2}}$                 ......(i)

This is a linear differential equation with

Integrating factor = $e^{\int\limits \frac{x}{x^2-1} d x}=e^{\frac{1}{2} \log \left|x^2-1\right|}=e^{\frac{1}{2} \log \left(1-x^2\right)}=\sqrt{1-x^2}$

Multiplying both sides of (i) by integrating factor and integrating with respect to $x$, we obtain

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

or, $y \sqrt{1-x^2}=\frac{x^5}{5}+x^2+C$               .....(ii)

Putting $x=0$ and $y=f(0)=0$ in (ii), we get $C=0$

Putting $C=0$ in (ii), we get

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

$\Rightarrow f(x)=\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}$

∴  $\int\limits_{-\sqrt{3} / 2}^{\sqrt{3} / 2} f(x) d x=\int\limits_{-\sqrt{3} / 2}^{\sqrt{3} / 2} \frac{x^5}{5 \sqrt{1-x^2}} d x+\int\limits_{-\sqrt{3} / 2}^{\sqrt{3} / 2} \frac{x^2}{\sqrt{1-x^2}} d x$

$=2 \int\limits_0^{\sqrt{3} / 2} \frac{x^2}{\sqrt{1-x^2}} d x$

$=2 \int\limits_0^{\pi / 3} \sin ^2 \theta d \theta$, where $x=\sin \theta$

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

$=\left[\theta-\frac{1}{2} \sin 2 \theta\right]_0^{\pi / 3}=\frac{\pi}{3}-\frac{\sqrt{3}}{4}$