Answer
$$ - \frac{b}{{{a^2}}}\ln \left| {\frac{x}{{a + bx}}} \right| - \frac{1}{{ax}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2}\left( {a + bx} \right)}}} dx \cr
& \frac{1}{{{x^2}\left( {a + bx} \right)}} = \frac{{Ax + B}}{{{x^2}}} + \frac{C}{{a + bx}} \cr
& \frac{1}{{{x^2}\left( {a + bx} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{a + bx}} \cr
& 1 = Ax\left( {a + bx} \right) + B\left( {a + bx} \right) + C{x^2} \cr
& 1 = Aax + Ab{x^2} + aB + Bbx + C{x^2} \cr
& 1 = Ab{x^2} + C{x^2} + Aax + Bbx + aB \cr
& Ab + C = 0 \cr
& Aa + Bb = 0 \cr
& aB = 1 \cr
& {\text{Solving the systems of equations simultaneously}} \cr
& B = \frac{1}{a} \cr
& Aa + Bb = 0 \to Aa + \frac{b}{a} = 0 \to A = - \frac{b}{{{a^2}}} \cr
& C = - Ab = \frac{{{b^2}}}{{{a^2}}} \cr
& {\text{Substituting}} \cr
& \frac{1}{{{x^2}\left( {a + bx} \right)}} = - \frac{b}{{{a^2}x}} + \frac{1}{{a{x^2}}} + \frac{{{b^2}}}{{{a^2}\left( {a + bx} \right)}} \cr
& {\text{Integrating}} \cr
& \int {\frac{1}{{{x^2}\left( {a + bx} \right)}}} dx = \int {\left[ { - \frac{b}{{{a^2}x}} + \frac{1}{{a{x^2}}} + \frac{{{b^2}}}{{{a^2}\left( {a + bx} \right)}}} \right]} dx \cr
& = - \frac{b}{{{a^2}}}\ln \left| x \right| - \frac{1}{{ax}} + \frac{b}{{{a^2}}}\ln \left| {a + bx} \right| + C \cr
& = - \frac{b}{{{a^2}}}\ln \left| {\frac{x}{{a + bx}}} \right| - \frac{1}{{ax}} + C \cr} $$
