What is the integral of x\sqrt[4]{3x-2} ?

The integral of x\sqrt[4]{3x-2} is 4/9 (1/9 (3x-2)^{9/4}+2/5 (3x-2)^{5/4})+C

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integral of x\sqrt[4]{3x-2}

\frac{4}{9} (\frac{1}{9} (3 x - 2)^{\frac{9}{4}} + \frac{2}{5} (3 x - 2)^{\frac{5}{4}}) + C

Solution steps

Apply u-substitution

: \int \frac{4 u ^{4} ( u ^{4} + 2 )}{9} d u

= \int \frac{4 u ^{4} ( u ^{4} + 2 )}{9} d u

Take the constant out:

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

= \frac{4}{9} \cdot \int u^{4} (u^{4} + 2) d u

Expand

u^{4} (u^{4} + 2) : u^{8} + 2 u^{4}

= \frac{4}{9} \cdot \int u^{8} + 2 u^{4} 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

= \frac{4}{9} (\int u^{8} d u + \int 2 u^{4} d u)

\int u^{8} d u = \frac{u ^{9}}{9}

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

= \frac{4}{9} (\frac{u ^{9}}{9} + \frac{2 u ^{5}}{5})

Substitute back

Simplify

= \frac{4}{9} (\frac{1}{9} (3 x - 2)^{\frac{9}{4}} + \frac{2}{5} (3 x - 2)^{\frac{5}{4}})

Add a constant to the solution

= \frac{4}{9} (\frac{1}{9} (3 x - 2)^{\frac{9}{4}} + \frac{2}{5} (3 x - 2)^{\frac{5}{4}}) + C

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What is the integral of x\sqrt[4]{3x-2} ?
The integral of x\sqrt[4]{3x-2} is 4/9 (1/9 (3x-2)^{9/4}+2/5 (3x-2)^{5/4})+C