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

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

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integral of (4x)/(\sqrt[3]{2x+1)}

\int \frac{4 x}{3 2 x + 1} d x

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

Solution steps

\int \frac{4 x}{3 2 x + 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{3 2 x + 1} d x

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \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}

= 4 \cdot \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

= 4 \cdot \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}}

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

Substitute back

u = 2 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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