Answer
$$\frac{1}{x} + 2\ln \left| {\frac{{x - 1}}{x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{x + 1}}{{{x^2}\left( {x - 1} \right)}}dx} \cr
& {\text{Decompose the integrand into partial fractions.}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{{x + 1}}{{{x^2}\left( {x - 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 1}} \cr
& \cr
& {\text{Multiplying by }}{x^2}\left( {x - 1} \right){\text{, we have}} \cr
& x + 1 = Ax\left( {x - 1} \right) + B\left( {x - 1} \right) + C{x^2} \cr
& {\text{expanding}} \cr
& x + 1 = A{x^2} - Ax + Bx - B + C{x^2} \cr
& {\text{group like terms}} \cr
& x + 1 = \left( {A{x^2} + C{x^2}} \right) + \left( { - Ax + Bx} \right) - B \cr
& {\text{comparing coefficients}}{\text{, we get the system}} \cr
& A + C = 0,\,\,\,\, - A + B = 1,\,\,\, - B = 1 \cr
& \cr
& {\text{Solving, we obtain}} \cr
& A = - 2,\,\,\,B = - 1,\,\,\,C = 2 \cr
& \cr
& {\text{The decomposition of the integrand is}} \cr
& \frac{{x + 1}}{{{x^2}\left( {x - 1} \right)}} = - \frac{2}{x} - \frac{1}{{{x^2}}} + \frac{2}{{x - 1}} \cr
& \cr
& \int {\frac{{x + 1}}{{{x^2}\left( {x - 1} \right)}}dx} = - \int {\frac{2}{x}dx - \int {\frac{1}{{{x^2}}}dx} + \int {\frac{2}{{x - 1}}} dx} \cr
& {\text{Integrating, we get:}} \cr
& = - 2\ln \left| x \right| + \frac{1}{x} + 2\ln \left| {x - 1} \right| + C \cr
& = \frac{1}{x} + 2\ln \left| {\frac{{x - 1}}{x}} \right| + C \cr} $$
