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

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

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integral of 1/(y^{1/3)}

\int \frac{1}{y ^{\frac{1}{3}}} d y

\frac{3}{2} y^{\frac{2}{3}} + C

Solution steps

\int \frac{1}{y ^{\frac{1}{3}}} d y

Apply the Power Rule

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

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

Add a constant to the solution

= \frac{3}{2} y^{\frac{2}{3}} + C

(x - 1) y^{^{'}} + (2 x + 1) y = 0

\frac{d}{d s} (ln (\frac{s - 1}{s ^{2} + 2 s + 5}))

f (x) = x^{2} - x + 2, at x = - 1

9 y^{^{''}} + 18 y^{^{'}} + 17 y = 0

n = 1 \sum \infty \frac{2 n}{5 n + 1}

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