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

\frac{\partial}{\partial x} (y^{2} - 4 x)

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

\int \frac{x ^{2} - 4}{4 x} d x

n = 0 \sum \infty \frac{n + 1}{2 n - 3}

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

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