How do you find the indefinite integral of #int x/(sqrt(9x^2))#?
2 Answers
By parts. Set
Explanation:
Then
The integration by parts formula is:
(or
so that
giving the integral shown in the answer.
Notes:

I assume that either you can quote the standard integral
#int 1/sqrt(a^2x^2)dx=sin^{1} (x/a)# (or#cos^{1}(x/a)# , it makes little difference). If not, you have to remember that you can differentiate all the inverse trig functions e.g.#y=sin^{1}x# by getting the#x# on to the left first:#x=sin y# , then differentiate with respect to#y# not#x# , getting#dx/{dy}=cos y = sqrt(1x^2)# , then reciprocate both sides#dy/{dx}=1/(sqrt(1x^2)# . Then the integration is reverse of differentiation. 
You have to guess correctly which part to make
#u(x)# and which to make#v(x)# . A simple rule of thumb here is that that the#x# on the top is an irritation and so if you differentiate it will turn into 1 and go away. If you guess wrong the#x# will turn into an#x^2# which is bad news. Similarly if the#x# had been and#x^2# you would have needed two rounds of integration by parts. 
I gloss over certain issues about "principal value" and the choice of
#sin^1x# or#cos^1x# .
Explanation:
#I=intx/sqrt(9x^2)dx#
Let
#I=1/2int(2x)/sqrt(9x^2)dx=1/2int1/sqrtudu=1/2intu^(1/2)du#
Using
#I=1/2(u^(1/2)/(1/2))=sqrtu=sqrt(9x^2)+C#