JEE Advanced 2014 · Paper 2, Mathematics question 41

Integrating factor. On $(-1,1)$,
$$\exp\int\frac{x}{x^2-1}dx=\exp\left(\tfrac12\ln(1-x^2)\right)=\sqrt{1-x^2}.$$

Solve.
$$\frac{d}{dx}\left(y\sqrt{1-x^2}\right)=x^4+2x\implies y\sqrt{1-x^2}=\frac{x^5}5+x^2+C,$$
and $f(0)=0$ gives $C=0$. So $f(x)=\dfrac{x^5/5+x^2}{\sqrt{1-x^2}}$.

Integrate over the symmetric interval. The $x^5$ term is odd and drops out. With $x=\sin\theta$:
$$\int_{-\sqrt3/2}^{\sqrt3/2}\frac{x^2}{\sqrt{1-x^2}}dx=2\int_0^{\pi/3}\sin^2\theta\,d\theta=\left[\theta-\tfrac12\sin2\theta\right]_0^{\pi/3}=\frac\pi3-\frac{\sqrt3}4.$$

(B)