Answer
$\frac{1}{a}\ln \left| {ax + b} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{ax + b}}} ,{\text{ }}a \ne 0 \cr
& {\text{Let }}u = ax + b,{\text{ then }}du = adx,{\text{ }}dx = \frac{1}{a}du,{\text{ then substituting}} \cr
& \int {\frac{{dx}}{{ax + b}}} = \int {\frac{1}{u}\left( {\frac{1}{a}} \right)du} \cr
& = \int {\frac{1}{u}\left( {\frac{1}{a}} \right)du} \cr
& = \frac{1}{a}\int {\frac{1}{u}du} \cr
& {\text{Integrate }} \cr
& = \frac{1}{a}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = ax + b \cr
& = \frac{1}{a}\ln \left| {ax + b} \right| + C \cr} $$
