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(B) The given differential equation is $\frac{dy}{dx} + \frac{x}{x^2-1}y = \frac{x^4+2x}{\sqrt{1-x^2}}$.This is a linear differential equation of the

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

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(B) The given differential equation is $\frac{dy}{dx} + \frac{x}{x^2-1}y = \frac{x^4+2x}{\sqrt{1-x^2}}$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{x}{x^2-1}$ and $Q(x) = \frac{x^4+2x}{\sqrt{1-x^2}}$.The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{x}{x^2-1} dx} = e^{\frac{1}{2} \ln|x^2-1|} = \sqrt{1-x^2}$ (since $x^2 The general solution is $y \cdot 	ext{I.F.} = \int Q(x) \cdot 	ext{I.F.} dx + C$.$y \sqrt{1-x^2} = \int \frac{x^4+2x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx = \int (x^4+2x) dx = \frac{x^5}{5} + x^2 + C$.Since the curve passes through the origin $(0,0)$,we have $0 = 0 + 0 + C$,so $C = 0$.Thus,$f(x) = \frac{x^5}{5\sqrt{1-x^2}} + \frac{x^2}{\sqrt{1-x^2}}$.We need to evaluate $I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx$.Since $f(x)$ is the sum of an odd function $\frac{x^5}{5\sqrt{1-x^2}}$ and an even function $\frac{x^2}{\sqrt{1-x^2}}$,the integral of the odd part over the symmetric interval is $0$.So,$I = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$.Let $x = \sin 	heta$,then $dx = \cos 	heta d	heta$. When $x=0, 	heta=0$; when $x=\frac{\sqrt{3}}{2}, 	heta=\frac{\pi}{3}$.$I = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 	heta}{\cos 	heta} \cos 	heta d	heta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2 	heta d	heta = \int_{0}^{\frac{\pi}{3}} (1 - \cos 2	heta) d	heta = [	heta - \frac{\sin 2	heta}{2}]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\sin(2\pi/3)}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

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

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