Answer
\[ = \frac{{2\sqrt {ax + b} }}{{{a^2}}}\,\,\left[ {\,\frac{{\,\left( {ax - 2b} \right)}}{3}} \right] + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} \hfill \\
\hfill \\
set\,\,\,{u^2} = ax + b\,\,\,\,then\,\,\,\,x = \frac{{{u^2} - b}}{a} \hfill \\
and \hfill \\
2udu = adx\,\,\, \to dx = \frac{{2udu}}{a} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} = \frac{1}{a}\int_{}^{} {\,\left( {\frac{{{u^2} - b}}{{au}}} \right)\,\left( {2u} \right)du} \hfill \\
\hfill \\
Simplify \hfill \\
\hfill \\
\int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} = \frac{2}{{{a^2}}}\int_{}^{} {\,\left( {{u^2} - b} \right)} du \hfill \\
\hfill \\
integrating \hfill \\
\hfill \\
\int_{}^{} {\frac{x}{{\sqrt {ax + b} }}dx} = \frac{2}{{{a^2}}}\,\,\left[ {\frac{{{u^3}}}{3} - bu} \right] + C \hfill \\
substitute\,\,back \hfill \\
\hfill \\
= \frac{{2\sqrt {ax + b} }}{{{a^2}}}\,\,\left[ {\,\frac{{\,\left( {ax - 2b} \right)}}{3}} \right] + C \hfill \\
\end{gathered} \]
