∫ 12x² / (2x + 1) dx = \na 2x³ln|2x + 1| + c\nb 4x³ln|2x + 1| + c\nc 3x² - 3x + 3/2ln|2x + 1| + c\nd 3x²…

Explanation:

Step1: Perform polynomial long - division

Divide $12x^{2}$ by $2x + 1$. We have $12x^{2}=(6x - 3)(2x + 1)+3$. So, $\frac{12x^{2}}{2x + 1}=6x-3+\frac{3}{2x + 1}$.

Step2: Integrate term - by - term

$\int(6x - 3+\frac{3}{2x + 1})dx=\int6xdx-\int3dx+\int\frac{3}{2x + 1}dx$. For $\int6xdx = 6\times\frac{x^{2}}{2}=3x^{2}$, for $\int3dx=3x$, and for $\int\frac{3}{2x + 1}dx$. Let $u = 2x+1$, then $du=2dx$ and $\int\frac{3}{2x + 1}dx=\frac{3}{2}\int\frac{du}{u}=\frac{3}{2}\ln|u|=\frac{3}{2}\ln|2x + 1|$.

Step3: Combine the results

$\int(6x - 3+\frac{3}{2x + 1})dx=3x^{2}-3x+\frac{3}{2}\ln|2x + 1|+C$.

Answer:

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