What is the integral of x/((x+6)^{2/3)} ?

The integral of x/((x+6)^{2/3)} is 3/4 (x+6)^{4/3}-18(x+6)^{1/3}+C

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integral of x/((x+6)^{2/3)}

\int \frac{x}{( x + 6 ) ^{\frac{2}{3}}} d x

\frac{3}{4} (x + 6)^{\frac{4}{3}} - 18 (x + 6)^{\frac{1}{3}} + C

Solution steps

\int \frac{x}{( x + 6 ) ^{\frac{2}{3}}} d x

Apply u-substitution

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

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

Expand

\frac{u - 6}{u ^{\frac{2}{3}}} : u^{\frac{1}{3}} - \frac{6}{u ^{\frac{2}{3}}}

= \int u^{\frac{1}{3}} - \frac{6}{u ^{\frac{2}{3}}} 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{1}{3}} d u - \int \frac{6}{u ^{\frac{2}{3}}} d u

\int u^{\frac{1}{3}} d u = \frac{3}{4} u^{\frac{4}{3}}

\int \frac{6}{u ^{\frac{2}{3}}} d u = 18 u^{\frac{1}{3}}

= \frac{3}{4} u^{\frac{4}{3}} - 18 u^{\frac{1}{3}}

Substitute back

u = x + 6

= \frac{3}{4} (x + 6)^{\frac{4}{3}} - 18 (x + 6)^{\frac{1}{3}}

Add a constant to the solution

= \frac{3}{4} (x + 6)^{\frac{4}{3}} - 18 (x + 6)^{\frac{1}{3}} + C

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What is the integral of x/((x+6)^{2/3)} ?
The integral of x/((x+6)^{2/3)} is 3/4 (x+6)^{4/3}-18(x+6)^{1/3}+C