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

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

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

Solution steps

assuming

(1 + 2 x) \geq 0

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

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{2}{3}}} d u

= \int \frac{1}{2 u ^{\frac{2}{3}}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{2}{3}}} d u

Apply the Power Rule

: 3 u^{\frac{1}{3}}

= \frac{1}{2} \cdot 3 u^{\frac{1}{3}}

Substitute back

u = 1 + 2 x

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

Simplify

Add a constant to the solution

\int_{- 3}^{3} - 9 - x^{2} d x

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

\int \frac{1 + x ^{8}}{x ^{13}} d x

1 - 6 x, at x = - 1

- \frac{2}{( s + 2 ) ^{2}}

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