$\int \frac{x}{x^2+x-12} d x$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{3}{7} \log |x-3|+\frac{4}{7} \log |x+4|+C$
$\int \frac{x}{x^2+x-12} d x$
$=\int\frac{x}{(x+4)(x-3)}dx$
$⇒\frac{x}{(x+4)(x-3)}=\frac{A}{x+4}+\frac{B}{x-3}$
$⇒x=A(x-3)+B(x+4)$
$⇒x=(A+B)x+(-3A+4B)$
$∴A=\frac{4}{7},B=\frac{3}{7}$