# How to solve the integral of y=1/(x-2)^1/3?

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### 3 Answers

We have to integrate y = 1 / (x-2)^1/3

Int [ y ] = Int [ 1 / (x-2)^1/3 dx]

Let x - 2 = t

dt/dx = 1 => dt = dx

Int [ 1 / (x-2)^1/3 dx]

=> Int [ 1 / t^1/3 dt]

=> Int [ t^(-1/3) dt]

=> [t^( -1/3 + 1)]/ ( -1/3 +1) + C

=> t^(2/3) / (2/3) + C

=> (3/2)*t^(2/3) + C

substitute t with x - 2

=> (3/2)*(x - 2)^(2/3) + C

**Therefore the integral of 1 / (x-2)^1/3 is (3/2)*(x - 2)^(2/3) + C.**

To find the integral of y = 1/(x-2)^(1/3).

Int y dx = Int (1/(x-2)^(1/3)}dx.

Int ydx = IntÂ t^(-1/3) dt, where x-2 = t, dx = dt.

Int y dx = {1/(-1/3) +1)} t^(-1/3 +1) +C

Int ydx = (3/2)*t^(2/3) +C.

Now replace t = x-2.

Int y dx = 3{(x-2)^(2/3)}/2) +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**