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

The integral of (3x^2)/(\sqrt[3]{2+3x)} is 3/40 (5x^2-4x+4)(3x+2)^{2/3}+C

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

\frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}} + C

Solution steps

Take the constant out:

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

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{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

\int 4 u^{\frac{2}{3}} d u = \frac{12}{5} u^{\frac{5}{3}}

= 3 \cdot \frac{1}{27} (\frac{3}{8} u^{\frac{8}{3}} - \frac{12}{5} u^{\frac{5}{3}} + 6 u^{\frac{2}{3}})

Substitute back

u = 2 + 3 x

= 3 \cdot \frac{1}{27} (\frac{3}{8} (2 + 3 x)^{\frac{8}{3}} - \frac{12}{5} (2 + 3 x)^{\frac{5}{3}} + 6 (2 + 3 x)^{\frac{2}{3}})

Simplify

3 \cdot \frac{1}{27} (\frac{3}{8} (2 + 3 x)^{\frac{8}{3}} - \frac{12}{5} (2 + 3 x)^{\frac{5}{3}} + 6 (2 + 3 x)^{\frac{2}{3}}) : \frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}}

= \frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{40} (5 x^{2} - 4 x + 4) (3 x + 2)^{\frac{2}{3}} + C

What is the integral of (3x^2)/(\sqrt[3]{2+3x)} ?
The integral of (3x^2)/(\sqrt[3]{2+3x)} is 3/40 (5x^2-4x+4)(3x+2)^{2/3}+C