Answer
$$2\ln \left| {\frac{5}{6}} \right| - \frac{4}{5}$$
Work Step by Step
$$\eqalign{
& \int_1^5 {\frac{{x - 1}}{{{x^2}\left( {x + 1} \right)}}} dx \cr
& {\text{Integrating }}\int {\frac{{x - 1}}{{{x^2}\left( {x + 1} \right)}}} dx \cr
& \frac{{x - 1}}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} \cr
& x - 1 = Ax\left( {x + 1} \right) + B\left( {x + 1} \right) + C{x^2} \cr
& x - 1 = A{x^2} + Ax + Bx + B + C{x^2} \cr
& x - 1 = \left( {A{x^2} + C{x^2}} \right) + \left( {Ax + Bx} \right) + B \cr
& {\text{Comparing coefficients}} \cr
& A + C = 0 \cr
& A + B = 1 \cr
& B = - 1 \cr
& {\text{Solving the system of equations simultaneously}} \cr
& B = - 1,{\text{ }}A = 2,{\text{ }}C = - 2 \cr
& \frac{{x - 1}}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} \cr
& \frac{{x - 1}}{{{x^2}\left( {x + 1} \right)}} = \frac{2}{x} - \frac{1}{{{x^2}}} - \frac{2}{{x + 1}} \cr
& \int_1^5 {\frac{{x - 1}}{{{x^2}\left( {x + 1} \right)}}} dx = \int_1^5 {\left( {\frac{2}{x} - \frac{1}{{{x^2}}} - \frac{2}{{x + 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {2\ln \left| x \right| + \frac{1}{x} - 2\ln \left| {x + 1} \right|} \right]_1^5 \cr
& {\text{ = }}\left[ {2\ln \left| {\frac{x}{{x + 1}}} \right| + \frac{1}{x}} \right]_1^5 \cr
& = \left[ {2\ln \left| {\frac{5}{{5 + 1}}} \right| + \frac{1}{5}} \right] - \left[ {2\ln \left| {\frac{1}{1}} \right| + \frac{1}{1}} \right] \cr
& = 2\ln \left| {\frac{5}{6}} \right| + \frac{1}{5} - 1 \cr
& = 2\ln \left| {\frac{5}{6}} \right| - \frac{4}{5} \cr} $$
