So I currently have
$\int \sqrt{3-2x} \: \: x \, \, \, dx$
And I wanted to know a good way to solve it. I know how to do it by parts, but I wanted to use a simpler substitution. So I start:
$ u = \sqrt{3-2x}$
$ du = -2\frac {1}{\sqrt{3-2x}} dx$
Then I proceed to substitute and I get
$-2 \int u \,\sqrt{3-2x}\,\, x\,\, du $
so
$-2 \int u^2 \,\, x\,\, du $
Know, here is my real question. I know that I can do something here, I can put x in terms of u, am I right? The thing is that I don't know how to do it... I start doing something like $ x = \sqrt{3-2x}$ but ofcourse that's nonsense. I'm guessing I'll get something like $(3-u)^2$ but I've no clue how to get there. I don't know how to put x in terms of u here.
Thanks a ton for your help!!