Answer
\[ = \frac{1}{{{a^2}}}\,\,\,\,\left[ {\,\left( {ax + b} \right) - b\ln \left| {ax + b} \right|} \right] + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{x}{{ax + b}}\,\,dx} \hfill \\
\hfill \\
set = ax + b\,\,\,then\,\,\,x = \frac{{u - b}}{a}{\text{ }}and\,\,du = adx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{x}{{ax + b}}\,\,dx} = \frac{1}{a}\int_{}^{} {\,\left( {\frac{{u - b}}{{au}}} \right)du} \hfill \\
\hfill \\
\int_{}^{} {\frac{x}{{ax + b}}\,\,dx} = \frac{1}{{{a^2}}}\int_{}^{} {\,\frac{{u - b}}{u}du} \hfill \\
\hfill \\
= \frac{1}{{{a^2}}}\int_{}^{} {\,\left( {1 - \frac{b}{u}} \right)du} \hfill \\
\hfill \\
= \frac{1}{{{a^2}}}\,\,\,\,\left[ {u - b\ln \left| u \right|} \right] + C \hfill \\
\hfill \\
substitute\,\,back \hfill \\
\hfill \\
= \frac{1}{{{a^2}}}\,\,\,\,\left[ {\,\left( {ax + b} \right) - b\ln \left| {ax + b} \right|} \right] + C \hfill \\
\end{gathered} \]