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)}

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

Solution steps

Apply u-substitution

Expand

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 = \frac{3}{5} u^{\frac{5}{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

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