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

f (x) = (x - 1) (x - 3)^{2}, at x = 4

(1 + 2 y) d x - (4 - x) d y = 0

f (x) = 13 x

\frac{d}{d x} (arcsin (7 x + 1))

\int (8 x^{3}) d x

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