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

The integral of (x^2)/(\sqrt[3]{1+2x)} is 1/8 (3/8 (1+2x)^{8/3}-6/5 (1+2x)^{5/3}+3/2 (1+2x)^{2/3})+C

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

\frac{1}{8} (\frac{3}{8} (1 + 2 x)^{\frac{8}{3}} - \frac{6}{5} (1 + 2 x)^{\frac{5}{3}} + \frac{3}{2} (1 + 2 x)^{\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{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

\int 2 u^{\frac{2}{3}} d u = \frac{6}{5} u^{\frac{5}{3}}

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

Substitute back

u = 1 + 2 x

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

Add a constant to the solution

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

\frac{d}{d x} ((x - 2)^{2})

2 (x y - y) d x - x d y = 0

\int \frac{1}{x ^{\frac{1}{2}}} d x

\int x \cdot ln (5 + x) d x

\int \frac{π}{2} - 5 x^{- 0.5} d x

What is the integral of (x^2)/(\sqrt[3]{1+2x)} ?
The integral of (x^2)/(\sqrt[3]{1+2x)} is 1/8 (3/8 (1+2x)^{8/3}-6/5 (1+2x)^{5/3}+3/2 (1+2x)^{2/3})+C