$\int \frac{x}{x^2+x-12} d x$ is equal to

$\frac{3}{7} \log |x-3|+\frac{4}{7} \log |x+4|+C$ $-\frac{3}{7} \log |x-3|+\frac{4}{7} \log |x+4|+C$ $\frac{4}{7} \log |x-3|+\frac{3}{7} \log |x+4|+C$ $\frac{4}{7} \log |x-3|-\frac{3}{7} \log |x+4|+C$

$\frac{3}{7} \log |x-3|+\frac{4}{7} \log |x+4|+C$

The correct answer is Option (1) → $\frac{3}{7} \log |x-3|+\frac{4}{7} \log |x+4|+C$