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

The integral of x/(\sqrt[3]{1-2x)} is 1/4 (3/5 (1-2x)^{5/3}-3/2 (1-2x)^{2/3})+C

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integral of x/(\sqrt[3]{1-2x)}

\int \frac{x}{3 1 - 2 x} d x

\frac{1}{4} (\frac{3}{5} (1 - 2 x)^{\frac{5}{3}} - \frac{3}{2} (1 - 2 x)^{\frac{2}{3}}) + C

Solution steps

\int \frac{x}{3 1 - 2 x} d x

Apply u-substitution

: \int \frac{u - 1}{4 3 u} d u

= \int \frac{u - 1}{4 3 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{3 u} d u

Expand

\frac{u - 1}{3 u} : u^{\frac{2}{3}} - \frac{1}{3 u}

= \frac{1}{4} \cdot \int u^{\frac{2}{3}} - \frac{1}{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

= \frac{1}{4} (\int u^{\frac{2}{3}} d u - \int \frac{1}{3 u} d u)

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

\int \frac{1}{3 u} d u = \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{4} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 1 - 2 x

= \frac{1}{4} (\frac{3}{5} (1 - 2 x)^{\frac{5}{3}} - \frac{3}{2} (1 - 2 x)^{\frac{2}{3}})

Add a constant to the solution

= \frac{1}{4} (\frac{3}{5} (1 - 2 x)^{\frac{5}{3}} - \frac{3}{2} (1 - 2 x)^{\frac{2}{3}}) + C

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What is the integral of x/(\sqrt[3]{1-2x)} ?
The integral of x/(\sqrt[3]{1-2x)} is 1/4 (3/5 (1-2x)^{5/3}-3/2 (1-2x)^{2/3})+C