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

$-\frac{1}{3} \tan^{-1} x + \frac{2}{3} \tan^{-1} \frac{x}{2} + C$

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

Consider $\frac{x^2}{(x^2+1)(x^2+4)}$ and put $x^2 = y$.

Then

$\frac{x^2}{(x^2+1)(x^2+4)} = \frac{y}{(y+1)(y+4)}$

Write

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

So that

$y = A(y+4) + B(y+1)$

Comparing coefficients of $y$ and constant terms on both sides, we get $A+B=1$ and $4A+B=0$, which give

$A = -\frac{1}{3} \text{ and } B = \frac{4}{3}$

Thus,

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

Therefore,

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

$= -\frac{1}{3} \tan^{-1} x + \frac{4}{3} \cdot \frac{1}{2} \tan^{-1} \frac{x}{2} + C$

$= -\frac{1}{3} \tan^{-1} x + \frac{2}{3} \tan^{-1} \frac{x}{2} + C$