Solve: $\int \frac{6x}{x^2 + 2} dx + \int \frac{4}{x(x^2 + 2)} dx$

$2 \ln |x(x^2 + 2)| + C$

The correct answer is Option (4) → $2 \ln |x(x^2 + 2)| + C$

$I = \int \frac{6x}{x^2 + 2} dx + \int \frac{4}{x(x^2 + 2)} dx$

First integral: $\int \frac{6x}{x^2 + 2} dx$

Using substitution $u = x^2 + 2, du = 2x dx$, the integral simplifies to:

$3 \ln |x^2 + 2| + C_1$

Second integral: Decompose $\frac{4}{x(x^2 + 2)}$ using partial fractions:

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

Integrate both parts:

$\int \frac{2}{x} dx = 2 \ln |x|$

$\int \frac{2x}{x^2 + 2} dx = \ln |x^2 + 2|$

Second integral becomes: $2 \ln |x| - \ln |x^2 + 2| + C_2$

Combine the two integrals:

$3 \ln |x^2 + 2| + (2 \ln |x| - \ln |x^2 + 2|)$

$= 2 \ln |x^2 + 2| + 2 \ln |x|$

$= 2 \ln (|x(x^2 + 2)|) + C$