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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

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

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