∫13−2xdx equals -2ln|3 - 2x|+c 2(3−2x)2 +c -12ln|3 - 2x|+c -2(3−2x)2 +c

Answer:

C. $-\frac{1}{2}\ln|3 - 2x|+C$

Explanation:

Step1: Use substitution

Let $u = 3-2x$, then $du=-2dx$, so $dx=-\frac{1}{2}du$.

Step2: Rewrite the integral

$\int\frac{1}{3 - 2x}dx=\int\frac{1}{u}\left(-\frac{1}{2}du\right)=-\frac{1}{2}\int\frac{1}{u}du$.

Step3: Integrate $\frac{1}{u}$

We know that $\int\frac{1}{u}du=\ln|u| + C$.

Step4: Substitute back $u$

Substituting $u = 3-2x$ back, we get $-\frac{1}{2}\ln|3 - 2x|+C$.