What is the integral of x/(\sqrt[3]{1-3x)} ?

The integral of x/(\sqrt[3]{1-3x)} is -1/10 (2x+1)(-3x+1)^{2/3}+C

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integral of x/(\sqrt[3]{1-3x)}

- \frac{1}{10} (2 x + 1) (- 3 x + 1)^{\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{1}{9} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 1 - 3 x

= \frac{1}{9} (\frac{3}{5} (1 - 3 x)^{\frac{5}{3}} - \frac{3}{2} (1 - 3 x)^{\frac{2}{3}})

Simplify

\frac{1}{9} (\frac{3}{5} (1 - 3 x)^{\frac{5}{3}} - \frac{3}{2} (1 - 3 x)^{\frac{2}{3}}) : - \frac{1}{10} (2 x + 1) (- 3 x + 1)^{\frac{2}{3}}

= - \frac{1}{10} (2 x + 1) (- 3 x + 1)^{\frac{2}{3}}

Add a constant to the solution

= - \frac{1}{10} (2 x + 1) (- 3 x + 1)^{\frac{2}{3}} + C

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What is the integral of x/(\sqrt[3]{1-3x)} ?
The integral of x/(\sqrt[3]{1-3x)} is -1/10 (2x+1)(-3x+1)^{2/3}+C