Answer
$$\frac{1}{a}\ln \left| {\frac{x}{{a + bx}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\left( {a + bx} \right)}}} dx \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{1}{{x\left( {a + bx} \right)}} = \frac{A}{x} + \frac{B}{{a + bx}} \cr
& 1 = A\left( {a + bx} \right) + Bx \cr
& x = 0 \to A = \frac{1}{a} \cr
& x = - \frac{a}{b} \to B = - \frac{b}{a} \cr
& \frac{1}{{x\left( {a + bx} \right)}} = \frac{{1/a}}{x} + \frac{{ - b/a}}{{a + bx}} \cr
& \int {\frac{1}{{x\left( {a + bx} \right)}}} dx = \int {\left( {\frac{{1/a}}{x} + \frac{{ - b/a}}{{a + bx}}} \right)} dx \cr
& = \frac{1}{a}\int {\frac{1}{x}} dx - \frac{1}{a}\int {\frac{b}{{a + bx}}} + C \cr
& {\text{Integrate}} \cr
& = \frac{1}{a}\ln \left| x \right| - \frac{1}{a}\ln \left| {a + bx} \right| + C \cr
& {\text{Factor}} \cr
& = \frac{1}{a}\left( {\ln \left| x \right| - \ln \left| {a + bx} \right|} \right) + C \cr
& {\text{Using logarithmic properties}} \cr
& {\text{ = }}\frac{1}{a}\ln \left| {\frac{x}{{a + bx}}} \right| + C \cr} $$
