Solution:
\( \dfrac{dy}{dx}+\dfrac{xy}{x^2-1}=\dfrac{x^4+2x}{\sqrt{1-x^2}}, \, x \in (-1,1) \)
\( \dfrac{dy}{dx}+P(x)y=Q(x) \)
I.F. \( =e^{\displaystyle\int P(x)dx}=e^{\displaystyle\int \left(\dfrac{x}{x^2-1}\right)dx}\)
\(=e^{+\dfrac{1}{2}\displaystyle\int \left(\dfrac{-2x}{1-x^2}\right)dx} \) put \( 1-x^2=t=-2x\,dx=dt \)
\( \Rightarrow e^{\dfrac{1}{2}\ln (1-x^2)}=e^{\ln \sqrt{1-x^2}}=\sqrt{1-x^2} \)
How \( y\cdot \sqrt{1-x^2}=\int \dfrac{(x^4+2x)}{\sqrt{1-x^2}}\times (\sqrt{1-x^2})\,dx \)
\( \Rightarrow y\cdot \sqrt{1-x^2}=\int \left(x^4+2x\right)dx \Rightarrow y \times \sqrt{1-x^2}=\dfrac{x^5}{5}+x^2+C \)
it passes \( (0,0) \)
\( 0\times 1=0+C \Rightarrow \boxed{C=0} \)
\( f(x)=\dfrac{x^5+5x^2}{5\sqrt{1-x^2}} \)

\( \dfrac{1}{5}\displaystyle\int\limits_{-\dfrac{\sqrt{3}}{2}}^{\dfrac{\sqrt{3}}{2}} \dfrac{x^5}{\sqrt{1-x^2}}dx+ \displaystyle\int\limits_{-\dfrac{\sqrt{3}}{2}}^{\dfrac{\sqrt{3}}{2}} \dfrac{x^2}{\sqrt{1-x^2}}dx \)
\( 0+2\displaystyle\int\limits_{0}^{\dfrac{\sqrt{3}}{2}} \dfrac{x^2}{\sqrt{1-x^2}}dx \)
put \( x=\sin\theta \Rightarrow dx=\cos\theta d\theta \)
\( 2\displaystyle\int\limits_{0}^{\dfrac{\pi}{3}} \dfrac{\sin^2\theta}{\cos\theta}\times \cos\theta d\theta \)
\( \displaystyle\int\limits_{0}^{\dfrac{\pi}{3}} (1-\cos 2\theta) d\theta \Rightarrow \left(\theta-\dfrac{\sin 2\theta}{2}\right)_{0}^{\dfrac{\pi}{3}} \)
\( \dfrac{\pi}{3}-\dfrac{1}{2}\times \sin\left(\dfrac{2\pi}{3}\right) \)
\( \dfrac{\pi}{3}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2} \Rightarrow \dfrac{\pi}{3}-\dfrac{\sqrt{3}}{4} \)

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