Which is the integral of this expression? u-substitution

Question

I tried u-substitution, but I don´t know exactly where is my mistake when I do the process:

$\sqrt{(3-2x)}x^2dx$

The answer is:

$-\frac{3}{4}(3-2x)\sqrt{(3-2x)}+-\frac{3}{10}(3-2x)^2\sqrt{(3-2x)}--\frac{1}{28}(3-2x)^3\sqrt{(3-2x)}+c$

Answer 1

You could try the following: $$ u = 3-2x$$ $$ du = -2dx$$ $$ x = \frac{3-u}{2}$$

You will get $$\sqrt{u}\frac{(3-u)^2}{4}\frac{du}{-2}$$ This is easy to integrate using the power law of integration after expanding $(3-u)^2$.