How do you find the indefinite integral of #int sqrtx/(sqrtx-3)#?

I got:

#x + 6sqrtx + 18ln|sqrtx - 3| + C#

Let #u = sqrtx#. Then, #du = 1/(2sqrtx)dx#, #2udu = dx#, and we have:

#=> 2int u^2/(u - 3)du#

By subtracting and adding #9/(u - 3)#, we take advantage of the difference of two squares, and obtain:

#=> 2int (u^2 - 9)/(u - 3) + 9/(u - 3)du#

#= 2int ((u + 3)cancel((u - 3)))/cancel(u - 3) + 9/(u - 3)du#

#= 2int u+3 + 9/(u - 3)du#

#= 2(u^2/2 + 3u + 9ln|u - 3|)#

#= u^2 + 6u + 18ln|u - 3|#

But since #u = sqrtx#, we un-substitute to get:

#=> color(blue)(int sqrtx/(sqrt(x) - 3)dx = x + 6sqrtx + 18ln|sqrtx - 3| + C)#

#I=intsqrtx/(sqrtx-3)dx#

We will use the substitution #u=sqrtx-3#. This also implies that #sqrtx=u+3#, which is a substitution we will use in the numerator.

Differentiating both sides of the substitution, we see that #du=1/(2sqrtx)dx#, which looks (and is) a little messy. However, we can modify the integral to force this #1/(2sqrtx)dx# term to appear to our benefit!

Since there's currently no #1/(2sqrtx)# term present, we can multiply the integrand by #2sqrtx1/(2sqrtx)# to have no net change.

#I=intsqrtx/(sqrtx-3)(2sqrtx)(1/(2sqrtx)dx)#

#I=2int(sqrtxsqrtx)/(sqrtx-3)(1/(2sqrtx)dx)#

Returning to our substitutions, which we can now make more easily: we have let #sqrtx-3=u#, and following from this #sqrtx=u+3# and #1/(2sqrtx)dx=du#.

#I=2int(u+3)^2/udu#

Expanding the squared term:

#I=2int(u^2+6u+9)/udu#

Dividing through:

#I=2int(u+6+9/u)du#

We can write this as #3# separate integrals (don't forget to distribute the #2#):

#I=2intu+12intdu+18int1/udu#

All of these are fairly common integrals:

#I=2(u^2/2)+12u+18lnabsu+C#

From #u=sqrtx-3# we see that:

#I=(sqrtx-3)^2+12(sqrtx-3)+18lnabs(sqrtx-3)+C#

Expanding the first two terms to simplify:

#I=(x-6sqrtx+9)+12sqrtx-36+18lnabs(sqrtx-3)+C#

The two constants can merge into #C#, the all-powerful constant of integration:

#I=x+6sqrtx+18lnabs(sqrtx-3)+C#