I'm trying to solve this:

integral of (y^2) / sqrt(4-3y)

And so I see that I need to do a u-substitution. So I set u = 4-3y, and du = -3.

So at this point I have:

integral of ( (y^2) / sqrt(u) ) du/-3

= (-1/3) integral of ( (y^2) / sqrt(u) ) du

Now for the y^2 up top, I figure I'll need to do 3y = 4-u, but I'm not sure how to implement this. When I check my answers with wolfram alpha, they do the same thing up this point, but they put a 9 in front of the sqrt(u).

Something like this:

(-1/3) integral of ( ((4-u)^2) / 9sqrt(u) ) du

And so I am sort of stuck right now, and I'm not sure how to get past this step. Can someone tell me where the 9 is from? Perhaps I'm overlooking some crucial algebraic step.