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

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

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

\int \frac{1}{( y - 1 ) ^{\frac{1}{3}}} d y

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

Solution steps

\int \frac{1}{( y - 1 ) ^{\frac{1}{3}}} d y

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 = y - 1

= \frac{3}{2} (y - 1)^{\frac{2}{3}}

Add a constant to the solution

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

\int_{0}^{5} 3 x d x

\int e^{x} \cdot x^{2} d x

- 2 x^{3} + 17 x^{2} - 19 x - 14, [0, x]

\int \frac{cos ^{3} ( 4 x )}{sin ^{4} ( 4 x )} d x

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

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