Answer
$$2\ln \left| {x - 2} \right| - \ln \left| x \right| - \frac{3}{{x - 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + 3x - 4}}{{{x^3} - 4{x^2} + 4x}}} dx \cr
& \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{{x^2} + 3x - 4}}{{{x^3} - 4{x^2} + 4x}} = \frac{{{x^2} + 3x - 4}}{{x\left( {{x^2} - 4x + 4} \right)}} = \frac{{{x^2} + 3x - 4}}{{x{{\left( {x - 2} \right)}^2}}} \cr
& \frac{{{x^2} + 3x - 4}}{{x{{\left( {x - 2} \right)}^2}}} = \frac{A}{x} + \frac{B}{{x - 2}} + \frac{C}{{{{\left( {x - 2} \right)}^2}}} \cr
& {x^2} + 3x - 4 = A{\left( {x - 2} \right)^2} + Bx\left( {x - 2} \right) + Cx \cr
& {x^2} + 3x - 4 = A\left( {{x^2} - 4x + 4} \right) + B{x^2} - 2Bx + Cx \cr
& {x^2} + 3x - 4 = A{x^2} - 4Ax + 4A + B{x^2} - 2Bx + Cx \cr
& {x^2} + 3x - 4 = \left( {A{x^2} + B{x^2}} \right) + \left( { - 4Ax - 2Bx + Cx} \right) + 4A \cr
& 4A = - 4 \to A = - 1 \cr
& A + B = 1 \to B = 2 \cr
& - 4A - 2B + C = 3 \to C = 3 \cr
& \cr
& \frac{{{x^2} + 3x - 4}}{{x{{\left( {x - 2} \right)}^2}}} = \frac{{ - 1}}{x} + \frac{2}{{x - 2}} + \frac{3}{{{{\left( {x - 2} \right)}^2}}} \cr
& \int {\frac{{{x^2} + 3x - 4}}{{{x^3} - 4{x^2} + 4x}}} dx = \int {\left( {\frac{{ - 1}}{x} + \frac{2}{{x - 2}} + \frac{3}{{{{\left( {x - 2} \right)}^2}}}} \right)} dx \cr
& {\text{Integrate}} \cr
& {\text{ = }} - \ln \left| x \right| + 2\ln \left| {x - 2} \right| - \frac{3}{{x - 2}} + C \cr
& {\text{Rearranging}} \cr
& = 2\ln \left| {x - 2} \right| - \ln \left| x \right| - \frac{3}{{x - 2}} + C \cr} $$
