Question

How do you find the indefinite integral of `int root3x/(root3x-1)`?

Answer 1

`(root3x-1)^3+(9(root3x-1)^2)/2+9(root3x-1)+3ln(abs(root3x-1))+C`

Explanation:

We have `int root3x/(root3x-1)dx`

Substitute `u=(root3x-1)`
`(du)/(dx)=x^(-2/3)/3`

`dx=3x^(2/3)du`

`int root3x/(root3x-1)(3x^(2/3))du=int(3x)/(root3x-1)du=int(3(u+1)^3)/udu=3int(u^3+3u^2+3u+1)/udu=int3u^2+9u+9+3/udu=u^3+(9u^2)/2+9u+3ln(abs(u))+C`

Resubstitute `u=root3x-1`:
`(root3x-1)^3+(9(root3x-1)^2)/2+9(root3x-1)+3ln(abs(root3x-1))+C`