Answer
$$\eqalign{
& {\text{Trapezoidal Rule}} \approx 6.2461 \cr
& {\text{Simpson's Rule}} \approx 6.4615 \cr
& {\text{Graphing utility}} \approx 6.4377 \cr} $$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{{8x}}{{{x^2} + 4}}} dx \cr
& {\text{For }}n = 4,{\text{ }}\Delta x = \frac{{b - a}}{n} = \frac{{4 - 0}}{4} = 1,{\text{ then,}} \cr
& {x_0} = 0,{\text{ }}{x_1} = 1,{\text{ }}{x_2} = 2,{\text{ }}{x_3} = 3,{\text{ }}{x_4} = 4 \cr
& \cr
& {\text{*Using the trapezoidal Rule }}\left( {{\text{THEOREM 4}}{\text{.17}}} \right) \cr
& \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{2n}}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + \cdots 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr
& \int_0^4 {\frac{{8x}}{{{x^2} + 4}}} dx = \frac{1}{2}\left[ {\frac{{8\left( 0 \right)}}{{{{\left( 0 \right)}^2} + 4}} + \frac{{16\left( 1 \right)}}{{{{\left( 1 \right)}^2} + 4}} + \frac{{16\left( 2 \right)}}{{{{\left( 2 \right)}^2} + 4}}} \right] \cr
& {\text{ }} + \frac{1}{2}\left[ {\frac{{16\left( 3 \right)}}{{{{\left( 3 \right)}^2} + 4}} + \frac{{8\left( 4 \right)}}{{{{\left( 4 \right)}^2} + 4}}} \right] \cr
& {\text{Simplifying}} \cr
& \int_0^4 {\frac{{8x}}{{{x^2} + 4}}} dx \approx 6.2461 \cr
& \cr
& {\text{*Using Simpson's Rule }}\left( {{\text{THEOREM 4}}{\text{.19}}} \right) \cr
& \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{3n}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + \cdots } \right. \cr
& \left. {{\text{ }} + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr
& \int_0^4 {\frac{{8x}}{{{x^2} + 4}}} dx = \frac{1}{3}\left[ {\frac{{8\left( 0 \right)}}{{{{\left( 0 \right)}^2} + 4}} + \frac{{32\left( 1 \right)}}{{{{\left( 1 \right)}^2} + 4}} + \frac{{16\left( 2 \right)}}{{{{\left( 2 \right)}^2} + 4}}} \right] \cr
& {\text{ }} + \frac{1}{3}\left[ {\frac{{32\left( 3 \right)}}{{{{\left( 3 \right)}^2} + 4}} + \frac{{8\left( 4 \right)}}{{{{\left( 4 \right)}^2} + 4}}} \right] \cr
& {\text{Simplifying}} \cr
& \int_1^5 {\frac{{12}}{x}} dx \approx 6.4615 \cr
& \cr
& {\text{Using a graphing utility we obtain}} \cr
& \int_0^4 {\frac{{8x}}{{{x^2} + 4}}} dx \approx 6.4377 \cr} $$
