Answer
$$9\ln \left| {x - 6} \right| + 7\ln \left| {x + 2} \right| + \frac{8}{{x + 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{16{x^2}}}{{\left( {x - 6} \right){{\left( {x + 2} \right)}^2}}}dx} \cr
& {\text{integrand }} \cr
& = \frac{{16{x^2}}}{{\left( {x - 6} \right){{\left( {x + 2} \right)}^2}}} \cr
& {\text{partial fractions}} \cr
& \frac{{16{x^2}}}{{\left( {x - 6} \right){{\left( {x + 2} \right)}^2}}} = \frac{A}{{x - 6}} + \frac{B}{{x + 2}} + \frac{C}{{{{\left( {x + 2} \right)}^2}}} \cr
& 16{x^2} = A{\left( {x + 2} \right)^2} + B\left( {x - 6} \right)\left( {x + 2} \right) + C\left( {x - 6} \right) \cr
& {\text{multiplying}} \cr
& 16{x^2} = A\left( {{x^2} + 4x + 4} \right) + B\left( {{x^2} - 4x - 12} \right) + C\left( {x - 6} \right) \cr
& 16{x^2} = A{x^2} + 4Ax + 4A + B{x^2} - 4Bx - 12B + Cx - 6C \cr
& 16{x^2} = \left( {A{x^2} + B{x^2}} \right) + \left( {4Ax - 4Bx + Cx} \right) + \left( {4A - 12B - 6C} \right) \cr
& {\text{by equating the coefficients}} \cr
& {x^2}:{\text{ }}A + B = 16 \cr
& x:{\text{ }}4A - 4B + C = 0 \cr
& {x^0}:{\text{ }}4A - 12B - 6C = 0 \cr
& {\text{Solving these equations}} \cr
& A = 9 \cr
& B = 7 \cr
& C = - 8 \cr
& {\text{substituting constants}} \cr
& \frac{{16{x^2}}}{{\left( {x - 6} \right){{\left( {x + 2} \right)}^2}}} = \frac{9}{{x - 6}} + \frac{7}{{x + 2}} - \frac{8}{{{{\left( {x + 2} \right)}^2}}} \cr
& \int {\frac{{16{x^2}}}{{\left( {x - 6} \right){{\left( {x + 2} \right)}^2}}}dx} = \int {\left( {\frac{9}{{x - 6}} + \frac{7}{{x + 2}} - \frac{8}{{{{\left( {x + 2} \right)}^2}}}} \right)dx} \cr
& {\text{integrating}} \cr
& = 9\ln \left| {x - 6} \right| + 7\ln \left| {x + 2} \right| + \frac{8}{{x + 2}} + C \cr} $$
