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Question

# Calculate. $\int{x(1+x)^{1/3}}dx$

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$\begin{gathered} \int x {\left( {1 + x} \right)^{1/3}}dx \\ \textcolor{#4257b2}{{\text{Set }}\boxed{{u^3} = 1 + x} \Rightarrow \boxed{x = {u^3} - 1} \Rightarrow \boxed{dx = 3{u^2}du}} \\ \textcolor{#4257b2}{{\text{Apply the substitution}}} \\ \int {\overbrace x^{{u^3} - 1}} \overbrace {{{\left( {1 + x} \right)}^{1/3}}}^u\overbrace {dx}^{3{u^2}du} = \int {\left( {{u^3} - 1} \right)} \left( u \right)\left( {3{u^2}} \right)du \\ \textcolor{#4257b2}{ {\text{Simplify the integrand}}} \\ \int {\left( {3{u^5} - 3{u^3}} \right)du} \\ \textcolor{#4257b2}{{\text{Integrate}}{\text{, apply }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C}} \\ 3\left( {\frac{{{u^6}}}{6}} \right) - 3\left( {\frac{{{u^4}}}{4}} \right) + C \\ \frac{{{u^6}}}{2} - \frac{{3{u^4}}}{4} + C \\ \textcolor{#4257b2}{ {\text{Back - substitute }}u = \sqrt[3]{{1 + x}}} \\ \frac{{{{\left( {\sqrt[3]{{1 + x}}} \right)}^6}}}{2} - \frac{{3{{\left( {\sqrt[3]{{1 + x}}} \right)}^4}}}{4} + C \\ \textcolor{#4257b2}{ {\text{Simplify}}} \\ \frac{{{{\left( {1 + x} \right)}^2}}}{2} - \frac{{3\left( {1 + x} \right)\sqrt[3]{{1 + x}}}}{4} + C \\ \end{gathered}$

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