The function $y = f \left(x\right)$ 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 $f\left(0\right)=0$. Then${\int }_{-\sqrt{3}/2}^{\sqrt{3}/2} f\left(x\right)dx$ is

$\frac{dy}{dx}+\frac{x}{x^2-1}y=\frac{x^4+2x}{\sqrt{1-x^2}}$

The$I.F$. of this differential equation is

$e^{\int \frac{x}{x^2-1}dx}=e^{-\int \frac{x}{1-x^2}dx}=e^{\frac{1}{2}\ln \left(1-x^2\right)}=\sqrt{1-x^2}$

$\therefore$ Solution is $y\sqrt{1-x^2}=\int \frac{x\left(x^3+2\right)}{\sqrt{1-x^2}}\sqrt{1-x^2}dx+\lambda$

$=\int \left(x^4+2x\right)dx+\lambda =\frac{x^5}{5}+x^2+\lambda$

As $y\left(0\right)=0\Rightarrow \lambda =0\Rightarrow y\sqrt{1-x^2}=\frac{x^5}{5}+x^2$

Now ${\int }_{-\sqrt{3}/2}^{\sqrt{3}/2} \frac{x^5+x^2}{\sqrt{1-x^2}}dx={\int }_{-\sqrt{3}/2}^{\sqrt{3}/2} \frac{x^2}{\sqrt{1-x^2}}dx$

(The other part is odd)

 $=2{\int }_0^{\sqrt{3}/2} \frac{x^2}{\sqrt{1-x^2}}dx$ Let $x=\sin \theta ,$ we get

$\begin{array}{l}I=2{\int }_0^{\pi /3} \frac{{\sin }^2\theta }{\cos \theta }\cos \theta d\theta =2{\int }_0^{\pi /3} {\sin }^2\theta d\theta \\ ={\int }_0^{\pi /3} (1-\cos 2\theta )d\theta =\theta -{\left.\frac{\sin 2\theta }{2}\right|}_0^{\pi /3}=\frac{\pi }{3}-\frac{\sqrt{3}}{4}\end{array}$