Answer
$$ \frac{1}{{{b^3}}}\left( {bu - 2a\ln \left| {a + bu} \right| - \frac{{{a^2}}}{{a + bu}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{u^2}}}{{{{\left( {a + bu} \right)}^2}}}} du \cr
& {\text{Expand the denominator}} \cr
& = \int {\frac{{{u^2}}}{{{a^2} + 2abu + {b^2}{u^2}}}} du \cr
& {\text{By long division}} \cr
& \frac{{{u^2}}}{{{a^2} + 2abu + {b^2}{u^2}}} = \frac{1}{{{b^2}}} + \frac{{ - \frac{{2a}}{b}u - \frac{{{a^2}}}{{{b^2}}}}}{{{a^2} + 2abu + {b^2}{u^2}}} \cr
& = \int {\left( {\frac{1}{{{b^2}}} - \frac{{\frac{{2a}}{b}u + \frac{{{a^2}}}{{{b^2}}}}}{{{{\left( {a + bu} \right)}^2}}}} \right)} du \cr
& {\text{Decompose }}\frac{{\frac{{2a}}{b}u + \frac{{{a^2}}}{{{b^2}}}}}{{{{\left( {a + bu} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{{\frac{{2a}}{b}u + \frac{{{a^2}}}{{{b^2}}}}}{{{{\left( {a + bu} \right)}^2}}} = \frac{A}{{a + bu}} + \frac{B}{{{{\left( {a + bu} \right)}^2}}} \cr
& \frac{{2a}}{b}u + \frac{{{a^2}}}{{{b^2}}} = A\left( {a + bu} \right) + B \cr
& \frac{{2a}}{b}u + \frac{{{a^2}}}{{{b^2}}} = Aa + Abu + B \cr
& \frac{{2a}}{b}u + \frac{{{a^2}}}{{{b^2}}} = Abu + \left( {Aa + B} \right) \cr
& {\text{Equating coefficients}} \cr
& Ab = \frac{{2a}}{b} \to A = \frac{{2a}}{{{b^2}}} \cr
& Aa + B = \frac{{{a^2}}}{{{b^2}}} \to B = \frac{{{a^2}}}{{{b^2}}} - \frac{{2{a^2}}}{{{b^2}}} = - \frac{{{a^2}}}{{{b^2}}} \cr
& {\text{Therefore,}} \cr
& \frac{{\frac{{2a}}{b}u + \frac{{{a^2}}}{{{b^2}}}}}{{{{\left( {a + bu} \right)}^2}}} = \frac{{2a}}{{{b^2}\left( {a + bu} \right)}} - \frac{{{a^2}}}{{{b^2}{{\left( {a + bu} \right)}^2}}} \cr
& = \int {\left( {\frac{1}{{{b^2}}} - \frac{{2a}}{{{b^2}\left( {a + bu} \right)}} + \frac{{{a^2}}}{{{b^2}{{\left( {a + bu} \right)}^2}}}} \right)} du \cr
& = \int {\frac{1}{{{b^2}}}} du - \int {\frac{{2a}}{{{b^2}\left( {a + bu} \right)}}du + \int {\frac{{{a^2}}}{{{b^2}{{\left( {a + bu} \right)}^2}}}du} } \cr
& = \frac{1}{{{b^2}}}\int {du} - \frac{{2a}}{{{b^2}}}\int {\frac{1}{{a + bu}}du + \frac{{{a^2}}}{{{b^2}}}\int {\frac{1}{{{{\left( {a + bu} \right)}^2}}}du} } \cr
& {\text{Integrating}} \cr
& = \frac{1}{{{b^2}}}u - \frac{{2a}}{{{b^3}}}\ln \left| {a + bu} \right| - \frac{{{a^2}}}{{{b^3}\left( {a + bu} \right)}} + C \cr
& {\text{Factoring}} \cr
& = \frac{1}{{{b^3}}}\left( {bu - 2a\ln \left| {a + bu} \right| - \frac{{{a^2}}}{{a + bu}}} \right) + C \cr} $$
