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