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

We need to solve the differential equation $$\dfrac{dy}{dx} + \dfrac{xy}{x^2 - 1} = \dfrac{x^4 + 2x}{\sqrt{1-x^2}}$$ with $$y(0) = 0$$, then compute $$\displaystyle\int_{-\sqrt{3}/2}^{\sqrt{3}/2} f(x)\,dx$$.

The equation is linear: $$\dfrac{dy}{dx} + P(x)y = Q(x)$$ where $$P(x) = \dfrac{x}{x^2-1}$$.

$$\int P(x)\,dx = \int \dfrac{x}{x^2-1}\,dx = \dfrac{1}{2}\ln|x^2-1|$$

For $$x \in (-1,1)$$: $$x^2-1 < 0$$, so $$|x^2-1| = 1-x^2$$.

$$\text{IF} = e^{\frac{1}{2}\ln(1-x^2)} = \sqrt{1-x^2}$$

$$\dfrac{d}{dx}\left[y\sqrt{1-x^2}\right] = \dfrac{(x^4+2x)\sqrt{1-x^2}}{\sqrt{1-x^2}} = x^4 + 2x$$

$$y\sqrt{1-x^2} = \int(x^4+2x)\,dx = \dfrac{x^5}{5} + x^2 + C$$

Using $$y(0) = 0$$: $$0 = 0 + 0 + C$$, so $$C = 0$$.

$$y = f(x) = \dfrac{x^5/5 + x^2}{\sqrt{1-x^2}}$$

$$\int_{-\sqrt{3}/2}^{\sqrt{3}/2} \dfrac{x^5/5 + x^2}{\sqrt{1-x^2}}\,dx$$

Since $$x^5/5$$ is odd and $$x^2$$ is even:

$$= 0 + 2\int_0^{\sqrt{3}/2} \dfrac{x^2}{\sqrt{1-x^2}}\,dx$$

$$2\int_0^{\pi/3} \dfrac{\sin^2\theta}{\cos\theta}\cdot\cos\theta\,d\theta = 2\int_0^{\pi/3}\sin^2\theta\,d\theta$$

$$= 2\int_0^{\pi/3}\dfrac{1-\cos 2\theta}{2}\,d\theta = \left[\theta - \dfrac{\sin 2\theta}{2}\right]_0^{\pi/3}$$

$$= \dfrac{\pi}{3} - \dfrac{\sin(2\pi/3)}{2} = \dfrac{\pi}{3} - \dfrac{\sqrt{3}/2}{2} = \dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}$$

The correct answer is Option B: $$\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}$$.