Calculus of a Single Variable, Chapter 5, 5.2, Section 5.2, Problem 30

Solving indefinite integral by u-substitution, we follow:
int f(g(x))*g'(x) = int f(u) *du where we let u = g(x) .
By following the instruction to let "u" be the denominator of the integral,
it means we let: u = root(3)(x) -1
Find the derivative of "u" which is du = 1/(3x^(2/3))dx
Then du =1/(3x^(2/3))dx can be rearrange into 3x^(2/3)du =dx .
Applying u-substitution using u =root(3)(x)-1 and 3x^(2/3)du =dx .
int root(3)(x)/(root(3)(x)-1) dx = int root(3)(x)/u*3x^(2/3)du
= int (x^(1/3)*3x^(2/3))/udu
=int (3x^(1/3+2/3))/udu
=3 int x/udu
Note: x^(1/3+2/3) = x^(3/3)
=x^1 or x
Algebraic techniques:
From u = root(3)(x)-1 , we can rearrange it into root(3)(x)=u+1 .
Raising both sides by a power 3:
(root(3)(x))^3 =(u+1)^3
x = (u+1)*(u+1)*(u+1)
By FOIL: (u+1)*(u+1) = u*u +u*1+1*u+1*1
= u^2+u+u+1
= u^2+2u+1
Then let (u+1)(u+1) = u^2 +2u +1 in (u+1)(u+1)(u+1) :
(u+1)(u+1)(u+1) = (u+1)*(u^2+2u+1)
Applying distributive property:
(u+1)(u^2+2u+1) = u *(u^2+2u+1) + 1*(u^2+2u+1)
= u^3 +2u^2+u +u^2+2u+1
=u^3+3u^2+3u+1
then x = (u+1)*(u+1)*(u+1) is the same as
x =u^3+3u^2+3u+1
Substitute x=u^3+3u^2+3u+1 in 3 int x/udu :
3 int x/udu = 3 int (u^3+3u^2+3u+1 )/u du
= 3int (u^3/u+(3u^2)/u+(3u)/u+1/u) du
=3int (u^2+3u+3+1/u) du
Evaluating each term in separate integral:
3 * [ int u^2 *du+ int 3u*du+int 3*du+ int 1/u du]
where:
int u^2 *du = u^3/3
int 3u*du =(3u^2)/2
int 3*du = 3u
int 1/u du= ln|u|
3 * [ int u^2 *du+ int 3u*du+int 3*du+ int 1/u du] becomes:
3*[u^3/3 +(3u^2)/2 +3u+ln|u|] +C= 3u^3/3 +(9u^2)/2 +9u+3ln|u|+C
Substitute u = root(3)(x)-1:
3u^3/3 +(9u^2)/2 +9u+3ln|u| +C = (root(3)(x)-1)^3 +(9(root(3)(x)-1)^2)/2 +9(root(3)(x)-1)+3ln|(root(3)(x)-1)| +C