Having a problem when figuring out a better way to solve $\int \sqrt{3-2x} \: \: x \, \, \, dx$ 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!!
 A: Since you are looking for a simpler way, here is one:
$$u= 3-2x \Rightarrow du =-2 dx, x= \frac{3-u}{2}$$
Then you get
$$\int \sqrt{3-2x} \: \: x \, \, \, dx = \frac{1}{-2} \int \sqrt{u} \frac{3-u}{2} du= -\frac{1}{4} \int 3 u^\frac{1}{2} -u^\frac{3}{2} du \,.$$
A: Hint 1: Rearrange the equation you have of u in terms of x ($u = \sqrt{3-2x}$) so that it gives you x in terms of u.
Hint 2: Also, check your work for when you found the derivative of u. Think again about how the power rule works. Fix this first and then think further about how you replaced du with dx. Make sure you rearrange your du equation to isolate dx and then plug that result into your integral. I think you have been a little careless in both these places.
A: To answer your question, note that $u^2=3-2x$ and therefore $x=\dfrac{3-u^2}{2}$.
Let's do the details. Your calculation of $du$ was almost fine, except for a missing $1/2$.  But I would rather write $u^2=3-2x$, and therefore $2u\,du=-2\,dx$. So $dx=-u\,du$.
Also, we have $x=\dfrac{3-u^2}{2}\,du$, so we want
$$\int (-u^2)\left(\frac{3-u^2}{2}\right)\,du.$$
Too many minus signs! We want to integrate $\frac{1}{2}(u^4-3u^2)$. Easy! 
