Answer
$$3\ln \left| x \right| - \ln \left| {x - 1} \right| - \frac{5}{{x - 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{x^2} + 3}}{{x{{\left( {x - 1} \right)}^2}}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{2{x^2} + 3}}{{x{{\left( {x - 1} \right)}^2}}} = \frac{A}{x} + \frac{B}{{x - 1}} + \frac{C}{{{{\left( {x - 1} \right)}^2}}} \cr
& 2{x^2} + 3 = A{\left( {x - 1} \right)^2} + Bx\left( {x - 1} \right) + Cx \cr
& 2{x^2} + 3 = A\left( {{x^2} - 2x + 1} \right) + B{x^2} - Bx + Cx \cr
& 2{x^2} + 3 = A{x^2} - 2Ax + A + B{x^2} - Bx + Cx \cr
& {\text{Collect like terms}} \cr
& 2{x^2} + 3 = \left( {A{x^2} + B{x^2}} \right) + \left( { - 2Ax - Bx + Cx} \right) + A \cr
& {\text{Solving we obtain}} \cr
& A = 3,{\text{ }}B = - 1,{\text{ }}C = 5 \cr
& {\text{Then}} \cr
& \frac{{2{x^2} + 3}}{{x{{\left( {x - 1} \right)}^2}}} = \frac{3}{x} - \frac{1}{{x - 1}} + \frac{5}{{{{\left( {x - 1} \right)}^2}}} \cr
& \int {\frac{{2{x^2} + 3}}{{x{{\left( {x - 1} \right)}^2}}}} dx = \int {\frac{3}{x}dx} - \int {\frac{1}{{x - 1}}dx + \int {\frac{5}{{{{\left( {x - 2} \right)}^2}}}dx} } \cr
& = 3\ln \left| x \right| - \ln \left| {x - 1} \right| - \frac{5}{{x - 2}} + C \cr} $$
