# IntegralSolve the integral of y=1/(x-2)^1/3.

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### 2 Answers

You need to evaluate the indefinite integral such that:

`int 1/(root(3) (x - 2))dx = int (x - 2)^(-1/3) dx`

`int (x - 2)^(-1/3) dx = ((x - 2)^(1 - 1/3))/(1 - 1/3) + c`

`int (x - 2)^(-1/3) dx = ((x - 2)^(2/3))/(2/3) + c`

`int (x - 2)^(-1/3) dx = (3/2)((x - 2)^(2/3)) + c`

**Hence, evaluating the indefinite integral yields **`int 1/(root(3) (x - 2))dx = (3/2)((x - 2)^(2/3)) + c.`

We'll use substitution technique to solve the integral.

We'll put x - 2 = t.

We'll differentiate both sides:

dx = dt

We'll re-write the integral in t:

Int dx/(x-2)^1/3 = Int dt/t^1/3

We'll use the negative power rule:

1/t^1/3 = t^(-1/3)

Int dt/t^1/3 = Int t^(-1/3)dt

Int t^(-1/3)dt = t^(-1/3 + 1)/(-1/3 + 1) + C

Int t^(-1/3)dt = t^(2/3)/(2/3) + C

**Int dx/(x-2)^1/3 = [3(x-2)^(2/3)]/2 + C**