Answer
$\frac{1}{5}\ln|\frac{x-4}{x+1}|+C$
Work Step by Step
$\frac{1}{x^{2}-3x-4}=\frac{1}{(x-4)(x+1)}$
$=\frac{A}{x-4}+\frac{B}{x+1}$
$⇒1= A(x+1)+B(x-4)$
Equating the coefficients of x and the constant term, we get A+B=0 and A-4B=1.
Solving these equations, we get $A= \frac{1}{5}$ and $B= -\frac{1}{5}$.
Thus $\frac{1}{x^{2}-3x-4}= \frac{1}{5(x-4)}-\frac{1}{5(x+1)}$
Therefore $\int\frac{dx}{x^{2}-3x-4}=\frac{1}{5}\int\frac{dx}{(x-4)}-\frac{1}{5}\int\frac{dx}{x+1}$
$=\frac{1}{5}(\ln|x-4|-\ln|x+1|)+C$
$=\frac{1}{5}\ln|\frac{x-4}{x+1}|+C$
