Answer
$$ - \frac{7}{2}\ln \left| {x - 1} \right| + \frac{9}{2}\ln \left| {x - 3} \right| + \frac{2}{{x - 3}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2} + x - 16}}{{\left( {x - 1} \right){{\left( {x - 3} \right)}^2}}}} dx \cr
& {\text{The partial fraction decomposition of the integrand is}} \cr
& \frac{{{x^2} + x - 16}}{{\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{Multiply the expression by }}\left( {x - 1} \right){\left( {x - 3} \right)^2} \cr
& {x^2} + x - 16 = A{\left( {x - 3} \right)^2} + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right) \cr
& {\text{If we set }}x = 1 \cr
& - 14 = A{\left( {1 - 3} \right)^2} \cr
& A = - \frac{7}{2} \cr
& \cr
& {\text{Simplify }}A{\left( {x - 3} \right)^2} + B\left( {x - 1} \right)\left( {x - 3} \right) + C\left( {x - 1} \right) \cr
& {x^2} + x - 16 = A\left( {{x^2} - 6x + 9} \right) + B\left( {{x^2} - 4x + 3} \right) + C\left( {x - 1} \right) \cr
& {x^2} + x - 16 = A{x^2} - 6Ax + 9A + B{x^2} - 4Bx + 3B + Cx - C \cr
& {\text{Collecting like powers of }}x \cr
& {x^2} + x - 16 = \left( {A{x^2} + B{x^2}} \right) + \left( { - 6Ax - 4Bx + Cx} \right) + \left( {9A + 3B - C} \right) \cr
& {x^2} + x - 16 = \left( {A + B} \right){x^2} + \left( { - 6A - 4B + C} \right)x + \left( {9A + 3B - C} \right) \cr
& {\text{We obtain the equations}} \cr
& A + B = 1,\,\,\, - 6A - 4B + C = 1,\,\,\,\,9A + 3B - C = - 16 \cr
& A + B = 1 \to - \frac{7}{2} + B = 1,\,\,B = \frac{9}{2} \cr
& - 6A - 4B + C = 1 \Rightarrow C = 1 - 6\left( { - \frac{7}{2}} \right) + 4\left( {\frac{9}{2}} \right) = - 2 \cr
& {\text{then}} \cr
& \frac{{{x^2} + x - 16}}{{\left( {x - 1} \right){{\left( {x - 3} \right)}^2}}} = \frac{{ - 7/2}}{{x - 1}} + \frac{{9/2}}{{x - 3}} + \frac{{ - 2}}{{{{\left( {x - 3} \right)}^2}}} \cr
& \cr
& \int {\frac{{{x^2} + x - 16}}{{\left( {x - 1} \right){{\left( {x - 3} \right)}^2}}}} dx = \int {\left( {\frac{{ - 7/2}}{{x - 1}} + \frac{{9/2}}{{x - 3}} + \frac{{ - 2}}{{{{\left( {x - 3} \right)}^2}}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = - \frac{7}{2}\ln \left| {x - 1} \right| + \frac{9}{2}\ln \left| {x - 3} \right| - 2\left( {\frac{{ - 1}}{{x - 3}}} \right) + C \cr
& = - \frac{7}{2}\ln \left| {x - 1} \right| + \frac{9}{2}\ln \left| {x - 3} \right| + \frac{2}{{x - 3}} + C \cr} $$
