Question

How do you integrate `(x^(1/3))/(((x^(1/3))-1))`?

Answer 1

`x+3/2x^(2/3)+3x^(1/3)+3lnabs(x^(1/3)-1)+C`

Explanation:

`I=intx^(1/3)/(x^(1/3)-1)dx`

Let `u=x^(1/3)-1`. This implies that `x^(1/3)=u+1` and that `x=(u+1)^3`, a fact we use to show that `dx=3(u+1)^2du`.

Substituting in what we know, this becomes:

`I=int(u+1)/u3(u+1)^2du=3int(u+1)^3/udu`

Expanding, then dividing:

`I=3int(u^3+3u^2+3u+1)/udu=3int(u^2+3u+3+1/u)du`

Integrating term by term:

`I=3(u^3/3+3/2u^2+3u+lnabsu)`

`I=u^3+9/2u^2+9u+3lnabsu`

Using `u=x^(1/3)-1`:

`I=(x^(1/3)-1)^3+9/2(x^(1/3)-1)^2+9(x^(1/3)-1)+3lnabs(x^(1/3)-1)`

If you wish, you can expand all of these terms and combine for the simplified answer of:

`I=(x-3x^(2/3)+3x^(1/3)-1)+9/2(x^(2/3)-2x^(1/3)+1)+9x^(1/3)-9+3lnabs(x^(1/3)-1)`

Continue with combining like terms, and have the `-9,1,` and `9/2` absorb into the constant of integration:

`I=x+3/2x^(2/3)+3x^(1/3)+3lnabs(x^(1/3)-1)+C`

Answer 2

I also got `x + 3x^"1/3" + 3/2x^"2/3" + 3ln|x^"1/3" - 1| + C`.

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Another way to do it is:

`int x^"1/3"/(x^"1/3" - 1)dx`

`= int cancel((x^"1/3" - 1)/(x^"1/3" - 1))^(1) + 1/(x^"1/3" - 1)dx`

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

For this, let `u = x^"1/3"` so that `du = 1/3x^(-"2/3")dx`, or `dx = 3x^"2/3"du = 3u^2du`. Then:

`=> 3int u^2du + 3int (u^2)/(u - 1)du`

Dividing the second integrand gives:

`((u^2 - 1) + 1)/(u - 1) = u + 1 + 1/(u - 1)`

so, this overall gives:

`=> 3int u^2du + 3int 1du + 3int udu + 3int 1/(u - 1)du`

`= u^3 + 3u + 3/2u^2 + 3ln|u - 1|`

`= color(blue)(x + 3x^"1/3" + 3/2x^"2/3" + 3ln|x^"1/3" - 1| + C)`

Answer 3

Making `y = x^(1/3)` we have `dy = 1/3x^(-2/3)dx = 1/3(dx)/y^2` so

`x^(1/3)/(x^(1/3)-1)dx = 3y^3/(y-1)dy = 3((y^3-1)/(y-1)+1/(y-1))dy=3(1+y+y^2+1/(y-1))dy`

so

`int x^(1/3)/(x^(1/3)-1)dx = 3int (1+y+y^2+1/(y-1))dy = 3y+3/2y^2+y^3+log(abs(y-1))+C` or

`int x^(1/3)/(x^(1/3)-1)dx=3x^(1/3)+3/2x^(2/3)+x + 3log(abs(x^(1/3)-1))+C`