Answer
$$
\int_{1}^{5} \frac{6}{2 x+1} d x
$$
Here $a=1, b=5,$ and $n=4,$ with $(b-a) / n=(5-1) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=1, x_{1}=2, x_{2}=3, x_{3}=4,$ and $x_{4}=5 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=5, a=1, f(x)=\frac{6}{2 x+1}\\
&\begin{array}{c|c|c}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 1 & 2 \\
1 & 2 & \frac{6}{5} \\
2 & 3 & \frac{6}{7} \\
3 & 4 & \frac{2}{3} \\
4 & 5 & \frac{6}{11} \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{aligned}
& \int_{1}^{5} \frac{6}{2 x+1} d x \\
& \quad \approx \frac{5-1}{4}\left[\frac{1}{2}(2)+\frac{6}{5}+\frac{6}{7}+\frac{2}{3}+\frac{1}{2}\left(\frac{6}{11}\right)\right] \\
& \quad = 1\left(1+\frac{6}{5}+\frac{6}{7}+\frac{2}{3}+\frac{3}{11}\right) \\
& \quad \approx 3.997
\end{aligned}
$$
(b) Simpson’s Rule:
$$
\begin{aligned}
& \int_{1}^{5} \frac{6}{2 x+1} d x \\
& \quad \approx \frac{5-1}{3(4)}\left[2+4\left(\frac{6}{5}\right)+2\left(\frac{6}{7}\right)+4\left(\frac{2}{3}\right)+\left(\frac{6}{11}\right)\right] \\
& \quad = \frac{1}{3}\left(2+\frac{24}{5}+\frac{12}{7}+\frac{8}{3}+\frac{6}{11}\right) \\
& \quad \approx 3.909
\end{aligned}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{1}^{5} \frac{6}{2 x+1} d x &=3 \ln \mid 2 x+1 \|_{1}^{5} \\
&=3(\ln |11|-\ln |3|) \\
&=3 \ln \frac{11}{3} \approx 3.898
\end{aligned}
$$
Work Step by Step
$$
\int_{1}^{5} \frac{6}{2 x+1} d x
$$
Here $a=1, b=5,$ and $n=4,$ with $(b-a) / n=(5-1) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=1, x_{1}=2, x_{2}=3, x_{3}=4,$ and $x_{4}=5 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=5, a=1, f(x)=\frac{6}{2 x+1}\\
&\begin{array}{c|c|c}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 1 & 2 \\
1 & 2 & \frac{6}{5} \\
2 & 3 & \frac{6}{7} \\
3 & 4 & \frac{2}{3} \\
4 & 5 & \frac{6}{11} \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{aligned}
& \int_{1}^{5} \frac{6}{2 x+1} d x \\
& \quad \approx \frac{5-1}{4}\left[\frac{1}{2}(2)+\frac{6}{5}+\frac{6}{7}+\frac{2}{3}+\frac{1}{2}\left(\frac{6}{11}\right)\right] \\
& \quad = 1\left(1+\frac{6}{5}+\frac{6}{7}+\frac{2}{3}+\frac{3}{11}\right) \\
& \quad \approx 3.997
\end{aligned}
$$
(b) Simpson’s Rule:
$$
\begin{aligned}
& \int_{1}^{5} \frac{6}{2 x+1} d x \\
& \quad \approx \frac{5-1}{3(4)}\left[2+4\left(\frac{6}{5}\right)+2\left(\frac{6}{7}\right)+4\left(\frac{2}{3}\right)+\left(\frac{6}{11}\right)\right] \\
& \quad = \frac{1}{3}\left(2+\frac{24}{5}+\frac{12}{7}+\frac{8}{3}+\frac{6}{11}\right) \\
& \quad \approx 3.909
\end{aligned}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{1}^{5} \frac{6}{2 x+1} d x &=3 \ln \mid 2 x+1 \|_{1}^{5} \\
&=3(\ln |11|-\ln |3|) \\
&=3 \ln \frac{11}{3} \approx 3.898
\end{aligned}
$$
