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

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{xy}}{{\mathrm{x}}^2-1}=\frac{{\mathrm{x}}^4+2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\\ \text{ I.F }={\mathrm{e}}^{\int \frac{\mathrm{x}}{{\mathrm{x}}^2-1}\mathrm{dx}}\\ \text{ I.F }=\sqrt{1-{\mathrm{x}}^2}\end{array}$

Solution of D.E.

$\begin{array}{l}\mathrm{y}\cdot \sqrt{1-{\mathrm{x}}^2}=\int \frac{{\mathrm{x}}^4+2\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}\cdot \sqrt{1-{\mathrm{x}}^2}\mathrm{dx}\\ \mathrm{y}\cdot \sqrt{1-{\mathrm{x}}^2}=\int \left({\mathrm{x}}^4+2\mathrm{x}\right)\mathrm{dx}\\ \mathrm{y}\cdot \sqrt{1-{\mathrm{x}}^2}=\frac{{\mathrm{x}}^5}{5}+{\mathrm{x}}^2+\mathrm{C}\\ \text{ At }\mathrm{x}=0,\mathrm{y}=0,\text{ get }\mathrm{C}=0\\ \mathrm{y}=\frac{{\mathrm{x}}^5}{5\sqrt{1-{\mathrm{x}}^2}}+\frac{{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\end{array}$

Now,

$\begin{array}{l}{\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{{\mathrm{x}}^5}{5\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}+{\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ {\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=0+2{\int }_0^{\frac{\sqrt{3}}{2}} \frac{{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^2}}\mathrm{dx}\\ {\int }_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{\pi }}{3}-\frac{\sqrt{3}}{4}\end{array}$