Answer
$$\frac{1}{5}\ln \left| {\frac{{x - 1}}{{x + 4}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2} + 3x - 4}}} \cr
& {\text{Factor the integrand}} \cr
& \frac{1}{{{x^2} + 3x - 4}} = \frac{1}{{\left( {x + 4} \right)\left( {x - 1} \right)}} \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{1}{{\left( {x + 4} \right)\left( {x - 1} \right)}} = \frac{A}{{x + 4}} + \frac{B}{{x - 1}} \cr
& {\text{Multiply the expression by }}\left( {x + 4} \right)\left( {x - 1} \right) \cr
& 1 = A\left( {x - 1} \right) + B\left( {x + 4} \right) \cr
& \cr
& {\text{If we set }}x = - 4 \cr
& 1 = A\left( { - 4 - 1} \right) + B\left( { - 4 + 4} \right),{\text{ }}A = - \frac{1}{5} \cr
& {\text{If we set }}x = 1 \cr
& 1 = A\left( {1 - 1} \right) + B\left( {1 + 4} \right),{\text{ }}B = \frac{1}{5} \cr
& {\text{Then}}{\text{,}} \cr
& \frac{A}{{x + 4}} + \frac{B}{{x - 1}} = \frac{{ - 1/5}}{{x + 4}} + \frac{{1/5}}{{x - 1}} \cr
& \int {\frac{{dx}}{{{x^2} + 3x - 4}}} = \int {\left( {\frac{{ - 1/5}}{{x + 4}} + \frac{{1/5}}{{x - 1}}} \right)} dx \cr
& = \frac{1}{5}\int {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 4}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{5}\left( {\ln \left| {x - 1} \right| - \ln \left| {x + 4} \right|} \right) + C \cr
& = \frac{1}{5}\ln \left| {\frac{{x - 1}}{{x + 4}}} \right| + C \cr} $$
