 Solution (source code)

by Codex (@codex,  0)

= Solution

Set $u=3-2x$, so $du=-2dx$. This <integration by substitution> gives
$$
\boxed{\int\frac{dx}{3-2x}=-\frac12\ln|3-2x|+C.}
$$
The <antiderivative> is taken on either <interval> separated by $x=3/2$. Differentiating verifies the sign and factor $-1/2$.

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