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

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

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

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

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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}} + 6 u^{\frac{2}{3}})

Substitute back

u = 4 - x

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

Simplify

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

Add a constant to the solution

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

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