Answer
$\ln \left| {\frac{{x - 1}}{{x + 4}}} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{5}{{\left( {x - 1} \right)\left( {x + 4} \right)}}} dx \cr
& {\text{Using the method of partial fractions}} \cr
& \frac{5}{{\left( {x - 1} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 4}} \cr
& {\text{Multiply the equation by }}\left( {x - 1} \right)\left( {x + 4} \right) \cr
& 5 = A\left( {x + 4} \right) + B\left( {x - 1} \right){\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Let }}x = 1{\text{ into }}\left( {\bf{1}} \right) \cr
& 5 = A\left( {1 + 4} \right) + B\left( {1 - 1} \right) \to A = 1 \cr
& {\text{Let }}x = - 4{\text{ into }}\left( {\bf{1}} \right) \cr
& 5 = A\left( { - 4 + 4} \right) + B\left( { - 4 - 1} \right) \to B = - 1 \cr
& {\text{Substituting }}A{\text{ and }}B{\text{ into the partial fraction decomposition}} \cr
& \frac{5}{{\left( {x - 1} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 4}} = \frac{1}{{x - 1}} - \frac{1}{{x + 4}} \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{5}{{\left( {x - 1} \right)\left( {x + 4} \right)}}} dx = \int {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 4}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \ln \left| {x - 1} \right| - \ln \left| {x + 4} \right| + C \cr
& {\text{Using logarithmic properties}} \cr
& = \ln \left| {\frac{{x - 1}}{{x + 4}}} \right| + C \cr} $$
