What is the integral of ((1+x^{1/3})^3)/(x^{1/2)} ?

The integral of ((1+x^{1/3})^3)/(x^{1/2)} is 2x^{1/2}+18/5 x^{5/6}+18/7 x^{7/6}+2/3 x^{3/2}+C

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integral of ((1+x^{1/3})^3)/(x^{1/2)}

\int \frac{( 1 + x ^{\frac{1}{3}} ) ^{3}}{x ^{\frac{1}{2}}} d x

2 x^{\frac{1}{2}} + \frac{18}{5} x^{\frac{5}{6}} + \frac{18}{7} x^{\frac{7}{6}} + \frac{2}{3} x^{\frac{3}{2}} + C

Solution steps

\int \frac{( 1 + x ^{\frac{1}{3}} ) ^{3}}{x ^{\frac{1}{2}}} d x

Expand

\frac{( 1 + x ^{\frac{1}{3}} ) ^{3}}{x ^{\frac{1}{2}}} : \frac{1}{x ^{\frac{1}{2}}} + \frac{3}{x ^{\frac{1}{6}}} + 3 x^{\frac{1}{6}} + x^{\frac{1}{2}}

= \int \frac{1}{x ^{\frac{1}{2}}} + \frac{3}{x ^{\frac{1}{6}}} + 3 x^{\frac{1}{6}} + x^{\frac{1}{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{x ^{\frac{1}{2}}} d x + \int \frac{3}{x ^{\frac{1}{6}}} d x + \int 3 x^{\frac{1}{6}} d x + \int x^{\frac{1}{2}} d x

\int \frac{1}{x ^{\frac{1}{2}}} d x = 2 x^{\frac{1}{2}}

\int \frac{3}{x ^{\frac{1}{6}}} d x = \frac{18}{5} x^{\frac{5}{6}}

\int 3 x^{\frac{1}{6}} d x = \frac{18}{7} x^{\frac{7}{6}}

\int x^{\frac{1}{2}} d x = \frac{2}{3} x^{\frac{3}{2}}

= 2 x^{\frac{1}{2}} + \frac{18}{5} x^{\frac{5}{6}} + \frac{18}{7} x^{\frac{7}{6}} + \frac{2}{3} x^{\frac{3}{2}}

Add a constant to the solution

= 2 x^{\frac{1}{2}} + \frac{18}{5} x^{\frac{5}{6}} + \frac{18}{7} x^{\frac{7}{6}} + \frac{2}{3} x^{\frac{3}{2}} + C

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What is the integral of ((1+x^{1/3})^3)/(x^{1/2)} ?
The integral of ((1+x^{1/3})^3)/(x^{1/2)} is 2x^{1/2}+18/5 x^{5/6}+18/7 x^{7/6}+2/3 x^{3/2}+C