The key here is to do a change of variables. Specifically, any time you see a square root, you either want to convert it to something squared under the square root or just set u= whatever is under it. In this case, we'll do the latter.

Let u=5x−1.Then, because we know \(\displaystyle{d}{u}=\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}{\left.{d}{x}\right.}\), we see that du=5dx which is equivalent to \(\displaystyle{k}{\left.{d}{x}\right.}=\frac{{1}}{{d}}{u}\). Substituting these into the integral, we get.

\(\displaystyle\int{x}\sqrt{{u}}\cdot\frac{{1}}{{5}}{d}{u}\)

This is part of what we want, but it still has that xx there and we want to totally convert to uu's. Well, remember that we set u=5x−1. Let's solve that for x in terms of u:

\(\displaystyle{5}{x}={u}+{1}\Rightarrow{x}=\frac{{{u}+{1}}}{{5}}\)

Substituting this in, we get

\(\displaystyle\int\frac{{{u}+{1}}}{{5}}\sqrt{{u}}\cdot\frac{{1}}{{5}}{d}{u}=\frac{{1}}{{25}}\int{\left({u}+{1}\right)}\sqrt{{u}}{d}{u}=\frac{{1}}{{25}}\int{\left({u}^{{\frac{{3}}{{2}}}}+{u}^{{\frac{{1}}{{2}}}}\right)}{d}{u}\)

Let u=5x−1.Then, because we know \(\displaystyle{d}{u}=\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}{\left.{d}{x}\right.}\), we see that du=5dx which is equivalent to \(\displaystyle{k}{\left.{d}{x}\right.}=\frac{{1}}{{d}}{u}\). Substituting these into the integral, we get.

\(\displaystyle\int{x}\sqrt{{u}}\cdot\frac{{1}}{{5}}{d}{u}\)

This is part of what we want, but it still has that xx there and we want to totally convert to uu's. Well, remember that we set u=5x−1. Let's solve that for x in terms of u:

\(\displaystyle{5}{x}={u}+{1}\Rightarrow{x}=\frac{{{u}+{1}}}{{5}}\)

Substituting this in, we get

\(\displaystyle\int\frac{{{u}+{1}}}{{5}}\sqrt{{u}}\cdot\frac{{1}}{{5}}{d}{u}=\frac{{1}}{{25}}\int{\left({u}+{1}\right)}\sqrt{{u}}{d}{u}=\frac{{1}}{{25}}\int{\left({u}^{{\frac{{3}}{{2}}}}+{u}^{{\frac{{1}}{{2}}}}\right)}{d}{u}\)