Calculus of a Single Variable, Chapter 5, 5.2, Section 5.2, Problem 29

To apply u-substitution , we let u = sqrt(x)-3 .
Then du = 1/(2sqrt(x) dx .
Rearrange du = 1/(2sqrt(x)) dx into dx =2sqrt(x) du
Substituting dx=2sqrt(x) du and u =sqrt(x)-3 :
int sqrt(x)/(sqrt(x)-3)dx = int sqrt(x)/u*2sqrt(x) dx
Simplify: sqrt(x)*sqrt(x) = x
int sqrt(x)/u *2sqrt(x) du = int (2x)/u du
Rearrange u=sqrt(x)-3 into sqrt(x)=u+3
Squaring both sides ofsqrt(x)=u+3 then
x=u^2+6u+9
int (2x)/u du = 2 int (u^2+6u+9)/u du
= 2 int (u^2/u + 6u/u + 9/u) du
= 2 int (u + 6 + 9/u) du
=2 *(u^2/2+6u+9lnabs|u|) +C
Substitute u =sqrt(x)-3:
2 *(u^2/2+6u+9ln|u|)+C =2 *((sqrt(x)-3)^2/2+6(sqrt(x)-3)+9ln|(sqrt(x)-3)|)+C
=(sqrt(x)-3)^2+12(sqrt(x)-3)+18ln|(sqrt(x)-3)| +C
= x-6sqrt(x)+9+12sqrt(x)-36 +18ln|sqrt(x)-3|+C
= x + 6sqrt(x)-27 +18ln|sqrt(x)-3|+C