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

3 Answers
Jan 7, 2017

#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#

Jan 8, 2017

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


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)#

Jan 8, 2017

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#