Take the integral:

\(\int{x}\sqrt{{{9}{x}^{2}+{4}}}{\left.{d}{x}\right.}\)

For the integrand x \(\sqrt{{{9}{x}^{2}+{4}}}\), substitute \({u}={9}{x}^{2}+{4}{\quad\text{and}\quad}{d}{u}={18}{x}{\left.{d}{x}\right.}\):

\(=\frac{1}{{18}}\int\sqrt{{{u}}}{d}{u}\)

The integral of \(\sqrt{{{u}}}\) is \(\frac{{{2}{u}^{{\frac{3}{{2}}}}}}{{3}}\):

\(=\frac{{u}^{{\frac{3}{{2}}}}}{{27}}+\text{constant}\)

Substitute back for \({u}={9}{x}^{2}+{4}\):

Answer:

\(=\frac{1}{{27}}{\left({9}{x}^{2}+{4}\right)}^{{\frac{3}{{2}}}}+\text{constant}\)