Eavaluating an definite integral Evaluate
$\int_{-1}^0  x^3 (1+x^3)^{\frac{1}{3}} dx$ 
I tried to solve it by parts as $ u=x  , dv=x^2(1+x^3)^{\frac{1}{3}}$
But I got another integral 
$\int_{-1}^0 (x^3 +1)^{\frac{4}{3}} dx $ and I could not evaluate it
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
\int_{-1}^{0}x^{3}\pars{1 + x^{3}}^{1/3}\,\dd x &
\,\,\,\stackrel{x\ \mapsto\ -x}{=}\,\,\,
-\int_{0}^{1}x^{3}\pars{1 - x^{3}}^{1/3}\,\dd x
\,\,\,\stackrel{x^{3}\ \mapsto\ x}{=}\,\,\,
-\,{1 \over 3}\int_{0}^{1}x^{1/3}\,\pars{1 - x}^{1/3}\,\dd x
\\[5mm] & =
-\,{1 \over 3}\,\mrm{B}\pars{{4 \over 3},{4 \over 3}}\qquad\pars{~\mrm{B}:\ Beta\ Function~}
\\[5mm] & =
-\,{1 \over 3}\,{\Gamma\pars{4/3}\Gamma\pars{4/3} \over
\Gamma\pars{4/3 + 4/3}}\qquad
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] & =
-\,{1 \over 3}\,{\bracks{\pars{1/3}\Gamma\pars{1/3}}^{\,2} \over
\pars{5/3}\pars{2/3}\Gamma\pars{2/3}} =
\bbx{-\,{1 \over 30}\,{\Gamma^{\,2}\pars{1/3} \over \Gamma\pars{2/3}}}
\approx -0.1767
\end{align}
