Answer
$ \ln |\frac{x-1}{x+4}|+C$
Work Step by Step
$\frac{5}{x^{2}+3x-4}=\frac{5}{(x+4)(x-1)}=\frac{A}{x+4}+\frac{B}{x-1}$
This gives $5= A(x-1)+B(x+4)$.
Putting x= 1 in the above equation, we get B=1.
Putting x= -4, we get A= -1.
Thus $\frac{5}{x^{2}+3x-4}= \frac{1}{x-1}-\frac{1}{x+4}$.
$\int \frac{5}{x^{2}+3x-4}= \int\frac{dx}{x-1}-\int\frac{dx}{x+4}$
$= \ln|x-1|- \ln|x+4| +C$
$= \ln |\frac{x-1}{x+4}|$+C
