Given the function
[tex]D\left(t\right)=\frac{6t}{1+2t}[/tex]Notice that
[tex]\begin{gathered} D\left(t\right)=6\left(\frac{t}{1+2t}\right)=6\left(\frac{1}{2}-\frac{1}{4t+2}\right) \\ \Rightarrow D\left(t\right)=6\left(\frac{1}{2}-\frac{1}{2\left(2t+1\right)}\right) \end{gathered}[/tex]Then, integrate D(t) from t=0 to t=3, as shown below
[tex]\Rightarrow\int_0^3D\left(t\right)dt=\int_0^36\left(\frac{1}{2}-\frac{1}{2\left(2t+1\right)}\right)dt=6\left(\frac{1}{2}\int_0^3dt-\frac{1}{2}\int_0^3\frac{dt}{\left(2t+1\right)}\right?[/tex]Then,
[tex]\Rightarrow\int_0^3D\left(t\right)dt=6\left(\frac{1}{2}*3-\frac{1}{2}\int_0^3\frac{dt}{2t+1}\right)[/tex]As for the remaining integral, set x=2t+1; then dx=2dt. Solving using that substitution,
as for the integration limits, when t=3, x=7, and when t=0, x=1.
[tex]\int_0^3\frac{dt}{2t+1}=\frac{1}{2}\int_1^7\frac{dx}{x}=\frac{1}{2}\left(ln\left(7\right)-ln\left(1\right)\right?=\frac{1}{2}\left(ln\left(\frac{7}{1}\right)\right)=\frac{1}{2}ln\left(7\right)[/tex]Therefore,
[tex]\Rightarrow\int_0^3D\left(t\right)dt=6\left(\frac{3}{2}-\frac{1}{2}\left(\frac{1}{2}ln\left(7\right)\right)\right)=9-\frac{3}{2}ln\left(7\right)[/tex]Rounding to three decimal places,
[tex]\Rightarrow\int_0^3D\lparen t)dt\approx6.081[/tex]Thus, the answer is 6.081
