What is the integral of (6x+1)/(\sqrt[3]{3x+2)} ?

The integral of (6x+1)/(\sqrt[3]{3x+2)} is 1/10 (12x-7)(3x+2)^{2/3}+C

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integral of (6x+1)/(\sqrt[3]{3x+2)}

\frac{1}{10} (12 x - 7) (3 x + 2)^{\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

Apply u-substitution

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 v^{\frac{2}{3}} d v = \frac{3}{5} v^{\frac{5}{3}}

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

Substitute back

= \frac{1}{3 \cdot 2 ^{\frac{2}{3}}} (\frac{3}{5} (6 x + 1 + 3)^{\frac{5}{3}} - \frac{9}{2} (6 x + 1 + 3)^{\frac{2}{3}})

Simplify

\frac{1}{3 \cdot 2 ^{\frac{2}{3}}} (\frac{3}{5} (6 x + 1 + 3)^{\frac{5}{3}} - \frac{9}{2} (6 x + 1 + 3)^{\frac{2}{3}}) : \frac{1}{10} (12 x - 7) (3 x + 2)^{\frac{2}{3}}

= \frac{1}{10} (12 x - 7) (3 x + 2)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{10} (12 x - 7) (3 x + 2)^{\frac{2}{3}} + C

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What is the integral of (6x+1)/(\sqrt[3]{3x+2)} ?
The integral of (6x+1)/(\sqrt[3]{3x+2)} is 1/10 (12x-7)(3x+2)^{2/3}+C