Given, $$\frac{d y}{d x}+\left(\frac{x}{x^2-1}\right) y=\frac{x^4+2 x}{\sqrt{1-x^2}}$$

The above differential equation is a linear differential equation.

Integrating factor I.F. $$=e^{\int \frac{x}{x^2-1} d x}$$

$$\begin{aligned} & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \int \frac{-2 x d x}{1-x^2}} \\ & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \ln \left(1-x^2\right)} \quad\left(\int \frac{f^{\prime}(x) d x}{f(x)}=\ln f(x)+c\right) \\ & \Rightarrow \text { I.F. }=e^{\ln \sqrt{1-x^2}} \\ & \Rightarrow \text { I.F. }=\sqrt{1-x^2} \end{aligned}$$

Now, the solution of the given differential equation is

$$\begin{aligned} & \Rightarrow y \cdot \text { (I.F. })=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right)(\text { L.F. }) d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right) \sqrt{1-x^2} d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\frac{x^5}{5}+2 \times \frac{x^2}{2}+\text { C } \\ & \Rightarrow y=f(x)=\frac{\frac{x^5}{5}+x^2+\mathrm{C}}{\sqrt{1-x^2}} \end{aligned}$$

Given, $$f(0)=0$$

$$\begin{array}{ll} \Rightarrow & 0=\frac{0+0+C}{\sqrt{1-0}} \\ \Rightarrow & C=0 \\ \therefore & f(x)=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}=\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}} \end{array}$$

Let $$\mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$$

$$\Rightarrow \mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\left(\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}\right) d x$$

Here, $$\frac{x^5}{5 \sqrt{1-x^2}}$$ is an odd function and $$\frac{x^2}{\sqrt{1-x^2}}$$ is an even function.

$$\begin{aligned} & \therefore \mathrm{I}=0+2 \int_0^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(\frac{x^2-1+1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(-\sqrt{1-x^2}+\frac{1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2\left[-\left(\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right)+\sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{\sqrt{3}}{4} \times \frac{1}{2}+\frac{1}{2} \times \frac{\pi}{3}\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi}{3}-\frac{\sqrt{3}}{4} \end{aligned}$$

Hint:

(i) The solution of linear differential equation $$\frac{d y}{d x}+\mathrm{P}(x) \cdot y=\mathrm{Q}(x)$$ is

$$y \text { (I.F.) } =\int \mathrm{Q}(x) \cdot(\text { I.F. }) d x+\text { C }$$

Where, $$\quad \text { L.F. } =e^{\int p(x) d x}$$

(ii) Recall the property

$$\int_{-a}^a f(x) d x=\left\{\begin{array}{l} 2 \int_0^a f(x) d x, \text { When } f(x) \text { is an even } \\ \text { function. } \\ 0, \text { When } f(x) \text { is an odd function. } \end{array}\right.$$