Find $\int \frac{x^2}{(x^2 + 1)(3x^2 + 4)} dx$.

$\frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{\sqrt{3}x}{2} \right) - \tan^{-1}x + C$

The correct answer is Option (2) → $\frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{\sqrt{3}x}{2} \right) - \tan^{-1}x + C$

Let $I = \int \frac{x^2}{(x^2 + 1)(3x^2 + 4)} dx$

Put $x^2 = t$

$\frac{t}{(t+1)(3t+4)} = \frac{A}{t+1} + \frac{B}{3t+4}$

$t = A(3t + 4) + B(t + 1)$

$t = (3A + B)t + (4A + B)$

On comparing both sides, we get:

$3A + B = 1 \text{ and } 4A + B = 0$

Solving these, we get $A = -1$ and $B = 4$.

$∴I = \int  \frac{-1}{x^2 + 1} + \frac{4}{3x^2 + 4}  dx$

$ = -\int \frac{1}{x^2 + 1} dx + 4 \int \frac{1}{3x^2 + 4} dx$

$ = -\int \frac{1}{x^2 + 1^2} dx + 4 \int \frac{1}{(\sqrt{3}x)^2 + 2^2} dx$

$ = -\frac{1}{1} \tan^{-1} x + \frac{4}{\sqrt{3}} \cdot \frac{1}{2} \tan^{-1} \left( \frac{\sqrt{3}x}{2} \right) + c$

$ = -\tan^{-1} x + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{3}x}{2} \right) + c$