What is the integral of (8x+2)^{1/3} ?

The integral of (8x+2)^{1/3} is 3/32 (8x+2)^{4/3}+C

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integral of (8x+2)^{1/3}

\int (8 x + 2)^{\frac{1}{3}} d x

\frac{3}{32} (8 x + 2)^{\frac{4}{3}} + C

Solution steps

\int (8 x + 2)^{\frac{1}{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{8} \cdot \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 8 x + 2

= \frac{1}{8} \cdot \frac{3}{4} (8 x + 2)^{\frac{4}{3}}

Simplify

\frac{1}{8} \cdot \frac{3}{4} (8 x + 2)^{\frac{4}{3}} : \frac{3}{32} (8 x + 2)^{\frac{4}{3}}

= \frac{3}{32} (8 x + 2)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{32} (8 x + 2)^{\frac{4}{3}} + C

\int cos (m y) d y

1 - 2 x

n = 1 \sum \infty \frac{2 n}{n !}

\int \frac{6 x ^{3} + 2 x ^{2} - 47 x - 9}{x ^{2} - 9} d x

f (x) = 7 x - 2 x^{2}, (- 1, - 9)

What is the integral of (8x+2)^{1/3} ?
The integral of (8x+2)^{1/3} is 3/32 (8x+2)^{4/3}+C