Answer
$\frac{1}{a}\ln \left| {\frac{{x - a}}{x}} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\left( {x - a} \right)}}} dx \cr
& {\text{Factoring the integrand}} \cr
& = \int {\frac{1}{{x\left( {x - a} \right)}}} dx \cr
& {\text{Using the method of partial fractions}} \cr
& \frac{1}{{x\left( {x - a} \right)}} = \frac{A}{x} + \frac{B}{{x - a}} \cr
& {\text{Multiply the equation by }}x\left( {x - 4} \right) \cr
& 1 = A\left( {x - a} \right) + Bx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Let }}x = 0{\text{ into }}\left( {\bf{1}} \right) \cr
& 1 = A\left( {0 - a} \right) + B\left( 0 \right){\text{ }} \to A = - \frac{1}{a} \cr
& {\text{Let }}x = a{\text{ into }}\left( {\bf{1}} \right) \cr
& 1 = A\left( {a - a} \right) + B\left( a \right) \to B = \frac{1}{a} \cr
& {\text{Substituting }}A{\text{ and }}B{\text{ into the partial fraction decomposition}} \cr
& \frac{1}{{x\left( {x - a} \right)}} = \frac{{ - 1/a}}{x} + \frac{{1/a}}{{x - a}} \cr
& {\text{Therefore}}{\text{,}} \cr
& \int {\frac{1}{{x\left( {x - a} \right)}}} dx = \int {\left( {\frac{{ - 1/a}}{x} + \frac{{1/a}}{{x - a}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = - \frac{1}{a}\ln \left| x \right| + \frac{1}{a}\ln \left| {x - a} \right| + C \cr
& {\text{Using logarithmic properties}} \cr
& = \frac{1}{a}\ln \left| {\frac{{x - a}}{x}} \right| + C \cr} $$
