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

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

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

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

Solution steps

Apply u-substitution

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

x \to 0 lim (1 + \frac{1}{x})

\int x e^{x^{3}} d x

\int 2 x - cos (6 x) d x

(y e^{- 2 x} + y^{3}) d x - e^{- 2 x} d y = 0

\int \frac{1}{- x ^{2} - x + 2} d x

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