Calculus of a Single Variable, Chapter 5, 5.2, Section 5.2, Problem 27

Solving for indefinite integral using u-substitution follows:
int f(g(x))*g'(x) dx = int f(u) du where we let u = g(x) .
In this case, it is stated that to let u be the denominator of integral which means let:
u = 1+sqrt(2x).
This can be rearrange into sqrt(2x) = u -1
Finding the derivative of u : du = 1/sqrt(2x) dx
Substituting sqrt(2x)= u-1 into du = 1/sqrt(2x)dx becomes:
du = 1/(u-1)dx
Rearranged into (u-1) du =dx
Applying u-substitution using u = 1+sqrt(2x) and (u-1)du = du :
int 1/(1+sqrt(2x)) dx = int (u-1)/u *du
Express into two separate fractions:
int (u-1)/u *du = int ( u/u -1/u)du
= int (1 - 1/u)du
Applying int (f(x) -g(x))dx = int f(x) dx - int g(x) dx :
int (1 - 1/u)du = int 1 du - int 1/udu
= u - ln|u| +C
Substitute u = 1+sqrt(2x) to the u - ln|u| +C :
u - ln|u| +C =1+sqrt(2x) -ln|1+sqrt(2x) |+C