What is the integral of x/(\sqrt[3]{x+6)} ?

The integral of x/(\sqrt[3]{x+6)} is 3/5 (x+6)^{5/3}-9(x+6)^{2/3}+C

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integral of x/(\sqrt[3]{x+6)}

\int \frac{x}{3 x + 6} d x

\frac{3}{5} (x + 6)^{\frac{5}{3}} - 9 (x + 6)^{\frac{2}{3}} + C

Solution steps

\int \frac{x}{3 x + 6} d x

Apply u-substitution

= \int \frac{u - 6}{3 u} d u

Expand

\frac{u - 6}{3 u} : u^{\frac{2}{3}} - \frac{6}{3 u}

= \int u^{\frac{2}{3}} - \frac{6}{3 u} d u

Apply the Sum Rule:

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

= \int u^{\frac{2}{3}} d u - \int \frac{6}{3 u} d u

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

\int \frac{6}{3 u} d u = 9 u^{\frac{2}{3}}

= \frac{3}{5} u^{\frac{5}{3}} - 9 u^{\frac{2}{3}}

Substitute back

u = x + 6

= \frac{3}{5} (x + 6)^{\frac{5}{3}} - 9 (x + 6)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{5} (x + 6)^{\frac{5}{3}} - 9 (x + 6)^{\frac{2}{3}} + C

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What is the integral of x/(\sqrt[3]{x+6)} ?
The integral of x/(\sqrt[3]{x+6)} is 3/5 (x+6)^{5/3}-9(x+6)^{2/3}+C