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

The integral of x\sqrt[3]{x-2} is 3/7 (x-2)^{7/3}+3/2 (x-2)^{4/3}+C

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integral of x\sqrt[3]{x-2}

\int x 3 x - 2 d x

\frac{3}{7} (x - 2)^{\frac{7}{3}} + \frac{3}{2} (x - 2)^{\frac{4}{3}} + C

Solution steps

\int x 3 x - 2 d x

Apply u-substitution

: \int (u + 2) 3 u d u

= \int (u + 2) 3 u d u

Expand

(u + 2) 3 u : u^{\frac{4}{3}} + 2 3 u

= \int u^{\frac{4}{3}} + 2 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

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

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

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

= \frac{3}{7} u^{\frac{7}{3}} + \frac{3}{2} u^{\frac{4}{3}}

Substitute back

u = x - 2

= \frac{3}{7} (x - 2)^{\frac{7}{3}} + \frac{3}{2} (x - 2)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (x - 2)^{\frac{7}{3}} + \frac{3}{2} (x - 2)^{\frac{4}{3}} + C

x \to 0 - lim (\frac{x ^{2}}{8} - \frac{4}{x})

\int_{2}^{4} 1 d x

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

\int_{0}^{6} 2 π x x^{2} + 9 d x

x^{3} + 8

What is the integral of x\sqrt[3]{x-2} ?
The integral of x\sqrt[3]{x-2} is 3/7 (x-2)^{7/3}+3/2 (x-2)^{4/3}+C