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

Solution

Verified

Step 1

1 of 2

\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}

Create a free account to view solutions

By signing up, you accept Quizlet's Terms of Service and Privacy Policy

By signing up, you accept Quizlet's Terms of Service and Privacy Policy

14th Edition•ISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir

10,142 solutions

10th Edition•ISBN: 9780471698043Einar Hille, Garret J. Etgen, Saturnino L. Salas

7,633 solutions

8th Edition•ISBN: 9781285741550James Stewart

11,083 solutions

9th Edition•ISBN: 9781337613927 (3 more)Daniel K. Clegg, James Stewart, Saleem Watson

11,050 solutions