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

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

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

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

$=\int\frac{x}{(x+4)(x-3)}dx$

$⇒\frac{x}{(x+4)(x-3)}=\frac{A}{x+4}+\frac{B}{x-3}$

$⇒x=A(x-3)+B(x+4)$

$⇒x=(A+B)x+(-3A+4B)$

$∴A=\frac{4}{7},B=\frac{3}{7}$