Answer
$\frac{1}{4}\ln \left| {4x + 7} \right| + C$
Work Step by Step
$$\eqalign{
& \text{Let } I=\int {\frac{{dx}}{{4x + 7}}} \cr
& {\text{Let }}u = 4x + 7,{\text{ then }}du = 4dx,{\text{ }}dx = \frac{1}{4}du \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& I = \int {\frac{1}{u}} \left( {\frac{1}{4}} \right)du \cr
& = \frac{1}{4}\int {\frac{1}{u}} du \cr
& {\text{Integrating}} \cr
& I= \frac{1}{4}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = 4x + 7 \cr
& I = \frac{1}{4}\ln \left| {4x + 7} \right| + C \cr} $$