What is the integral of (3x-1)^{1/8} ?

The integral of (3x-1)^{1/8} is 8/27 (3x-1)^{9/8}+C

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integral of (3x-1)^{1/8}

\int (3 x - 1)^{\frac{1}{8}} d x

\frac{8}{27} (3 x - 1)^{\frac{9}{8}} + C

Solution steps

\int (3 x - 1)^{\frac{1}{8}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{8}}}{3} d u

= \int \frac{u ^{\frac{1}{8}}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{\frac{1}{8}} d u

Apply the Power Rule

: \frac{8}{9} u^{\frac{9}{8}}

= \frac{1}{3} \cdot \frac{8}{9} u^{\frac{9}{8}}

Substitute back

u = 3 x - 1

= \frac{1}{3} \cdot \frac{8}{9} (3 x - 1)^{\frac{9}{8}}

Simplify

\frac{1}{3} \cdot \frac{8}{9} (3 x - 1)^{\frac{9}{8}} : \frac{8}{27} (3 x - 1)^{\frac{9}{8}}

= \frac{8}{27} (3 x - 1)^{\frac{9}{8}}

Add a constant to the solution

= \frac{8}{27} (3 x - 1)^{\frac{9}{8}} + C

\frac{\partial}{\partial x} (e^{y} + y e^{x})

\frac{d}{d x} (\frac{x + 3}{x - 4})

y = \frac{9 x ^{2} + 8}{x ^{2} + 1}

\int (\frac{6 x ^{5} - 3 x ^{2} + 4}{x ^{2}}) d x

\frac{d}{d x} (e^{s i n (x)})

What is the integral of (3x-1)^{1/8} ?
The integral of (3x-1)^{1/8} is 8/27 (3x-1)^{9/8}+C