How do you integrate $\int x(x+3)^ \left( 1/3 \right) dx$ I'm trying to do the following integral:
$\int x(x+3)^ \left( 1/3 \right) dx$
It is taken from this website. According to my source, it is to be done by substitution method. However, when I try that (using $u=x+3$), I have an $x$ term floating around, which is making things difficult. 
Could someone explain how this is done? (I'm fairly confident it can't be done by substitution). Thanks.
 A: let $u = x+3 \Rightarrow du = 1 dx \Rightarrow \int (u-3)(u)^{\frac{1}{3}} du.$
Now distribute and solve and change back to $x$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int x\pars{x + 3}^{1/3}\,\dd x:\ {\large ?}}$

\begin{align}
\color{#00f}{\large\int x\pars{x + 3}^{1/3}\,\dd x}&=
\left.{3 \over 4}\,\totald{}{\mu}\int\pars{\mu x + 3}^{4/3}\,\dd x
\right\vert_{\mu = 1}
=
\left.{9 \over 28}\,\totald{}{\mu}{\pars{\mu x + 3}^{7/3} \over \mu}
\right\vert_{\mu = 1}
\\[3mm]&={9 \over 28}\,\bracks{%
{7 \over 3}\,{\pars{\mu x + 3}^{4/3}x \over \mu}
-{\pars{\mu x + 3}^{7/3} \over \mu^{2}}
}_{\mu = 1}
\\[3mm]&={9 \over 28}\,\bracks{%
{7 \over 3}\,\pars{x + 3}^{4/3}x - {\pars{x + 3}^{7/3}}}
={9 \over 28}\,\pars{x + 3}^{4/3}\pars{{7 \over 3}\,x - x - 3}
\\[3mm]&=\color{#00f}{\large{3 \over 28}\,\pars{x + 3}^{4/3}\pars{4x - 9}}
+ \mbox{a constant}
\end{align}

A: Let $u=x+3$, then, $x=u-3$. You are done. 
$\int x(x+3)^{1/3}= \int (u-3)u^{1/3}.du $
