= Solution
Use <integration by substitution> with $u=3-2x$, so $dx=-du/2$ and $x=(3-u)/2$. The <indefinite integral> becomes
$$
-\frac14\int(3u^{1/2}-u^{3/2})\,du=-\frac12u^{3/2}+\frac1{10}u^{5/2}+C.
$$
Thus, with an arbitrary <constant of integration>,
$$
\boxed{\int x\sqrt{3-2x}\,dx=-\frac{x+1}{5}(3-2x)^{3/2}+C.}
$$
This real <antiderivative> applies for $x<3/2$ and extends continuously to the stated endpoint. Its left <derivative> there agrees with the limiting integrand, zero.
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