What is the integral of x/(\sqrt[3]{x-4)} ?

The integral of x/(\sqrt[3]{x-4)} is 3/5 (x-4)^{5/3}+6(x-4)^{2/3}+C

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integral of x/(\sqrt[3]{x-4)}

\int \frac{x}{3 x - 4} d x

\frac{3}{5} (x - 4)^{\frac{5}{3}} + 6 (x - 4)^{\frac{2}{3}} + C

Solution steps

\int \frac{x}{3 x - 4} d x

Apply u-substitution

: \int \frac{u + 4}{3 u} d u

= \int \frac{u + 4}{3 u} d u

Expand

\frac{u + 4}{3 u} : u^{\frac{2}{3}} + \frac{4}{3 u}

= \int u^{\frac{2}{3}} + \frac{4}{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{4}{3 u} d u

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

\int \frac{4}{3 u} d u = 6 u^{\frac{2}{3}}

= \frac{3}{5} u^{\frac{5}{3}} + 6 u^{\frac{2}{3}}

Substitute back

u = x - 4

= \frac{3}{5} (x - 4)^{\frac{5}{3}} + 6 (x - 4)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{5} (x - 4)^{\frac{5}{3}} + 6 (x - 4)^{\frac{2}{3}} + C

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What is the integral of x/(\sqrt[3]{x-4)} ?
The integral of x/(\sqrt[3]{x-4)} is 3/5 (x-4)^{5/3}+6(x-4)^{2/3}+C