Answer
$$\frac{1}{{{b^2}}}\left( {\ln \left| {a + bx} \right| + \frac{a}{{\left( {a + bx} \right)}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{{\left( {a + bx} \right)}^2}}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{x}{{{{\left( {a + bx} \right)}^2}}} = \frac{A}{{a + bx}} + \frac{B}{{{{\left( {a + bx} \right)}^2}}} \cr
& x = A\left( {a + bx} \right) + B \cr
& x = Aa + Abx + B \cr
& x = Abx + B + Aa \cr
& Ab = 1, \to A = \frac{1}{b} \cr
& B + Aa = 0 \to B = - Aa = - \frac{a}{b} \cr
& {\text{Substituting}} \cr
& \frac{x}{{{{\left( {a + bx} \right)}^2}}} = \frac{{1/b}}{{a + bx}} + \frac{{ - a/b}}{{{{\left( {a + bx} \right)}^2}}} \cr
& \int {\frac{x}{{{{\left( {a + bx} \right)}^2}}}} dx = \int {\left[ {\frac{{1/b}}{{a + bx}} + \frac{{ - a/b}}{{{{\left( {a + bx} \right)}^2}}}} \right]} dx \cr
& = \frac{1}{b}\int {\frac{1}{{a + bx}}} dx - \frac{a}{b}\int {\frac{1}{{{{\left( {a + bx} \right)}^2}}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{{{b^2}}}\ln \left| {a + bx} \right| - \frac{a}{{{b^2}}}\left( { - \frac{1}{{a + bx}}} \right) + C \cr
& = \frac{1}{{{b^2}}}\ln \left| {a + bx} \right| + \frac{a}{{{b^2}\left( {a + bx} \right)}} + C \cr
& {\text{Factoring}} \cr
& = \frac{1}{{{b^2}}}\left( {\ln \left| {a + bx} \right| + \frac{a}{{\left( {a + bx} \right)}}} \right) + C \cr} $$
