Question

enter an exact answer. provide your answer below. ∫₃² 3/(2x + 1) dx

Explanation:

Step1: Split the integral

Separate $\int_{2}^{3}\frac{3}{2x + 1}dx$ into $3\int_{2}^{3}\frac{1}{2x+1}dx$.

Step2: Use substitution

Let $u = 2x + 1$, then $du=2dx$, and $dx=\frac{1}{2}du$. When $x = 2$, $u=2\times2 + 1=5$; when $x = 3$, $u=2\times3+1 = 7$. So the integral becomes $\frac{3}{2}\int_{5}^{7}\frac{1}{u}du$.

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

The antiderivative of $\frac{1}{u}$ is $\ln|u|$. So $\frac{3}{2}\int_{5}^{7}\frac{1}{u}du=\frac{3}{2}[\ln(u)]_{5}^{7}$.

Step4: Evaluate the definite - integral

$\frac{3}{2}(\ln(7)-\ln(5))=\frac{3}{2}\ln(\frac{7}{5})$.

Answer:

$\frac{3}{2}\ln(\frac{7}{5})$