Answer
$$\ln \left| {x + 1} \right| + \ln \left| {x - 3} \right| + \frac{2}{{x - 3}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^2} - 10x + 4}}{{\left( {x + 1} \right){{\left( {x - 3} \right)}^2}}}} dx \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{{2{x^2} - 10x + 4}}{{\left( {x + 1} \right){{\left( {x - 3} \right)}^2}}} = \frac{A}{{x + 1}} + \frac{B}{{x - 3}} + \frac{C}{{{{\left( {x - 3} \right)}^2}}} \cr
& {\text{Multiplying by }}\left( {x + 1} \right){\left( {x - 3} \right)^2}{\text{ yields}} \cr
& 2{x^2} - 10x + 4 = A{\left( {x - 3} \right)^2} + B\left( {x + 1} \right)\left( {x - 3} \right) + C\left( {x + 1} \right) \cr
& 2{x^2} - 10x + 4 = A\left( {{x^2} - 6x + 9} \right) + B\left( {{x^2} - 2x - 3} \right) + Cx + C \cr
& 2{x^2} - 10x + 4 = A{x^2} - 6Ax + 9A + B{x^2} - 2Bx - 3B + Cx + C \cr
& {\text{Collecting like powers of }}x{\text{}} \cr
& 2{x^2} - 10x + 4 = \left( {A{x^2} + B{x^2}} \right) + \left( { - 6Ax - 2Bx + Cx} \right) + \left( {9A - 3B + C} \right) \cr
& {\text{Equating corresponding coefficients yields the following system }} \cr
& {\text{of linear equations}} \cr
& \,\,\,\,\,A + B = 2 \cr
& - 6A - 2B + C = - 10 \cr
& \,\,\,\,\,9A - 3B + C = 4 \cr
& {\text{Solving the system of linear equations we obtain}} \cr
& \,\,\,A = 1,\,\,\,B = 1,\,\,\,C = - 2 \cr
& \cr
& {\text{Then}}{\text{, the integrand can be written as}} \cr
& \,\,\frac{{2{x^2} - 10x + 4}}{{\left( {x + 1} \right){{\left( {x - 3} \right)}^2}}} = \frac{1}{{x + 1}} + \frac{1}{{x - 3}} - \frac{2}{{{{\left( {x - 3} \right)}^2}}} \cr
& \int {\frac{{2{x^2} - 10x + 4}}{{\left( {x + 1} \right){{\left( {x - 3} \right)}^2}}}} dx = \int {\left[ {\frac{1}{{x + 1}} + \frac{1}{{x - 3}} - \frac{2}{{{{\left( {x - 3} \right)}^2}}}} \right]} dx \cr
& {\text{Integrating}} \cr
& = \ln \left| {x + 1} \right| + \ln \left| {x - 3} \right| + \frac{2}{{x - 3}} + C \cr} $$
