[tex]\begin{cases}x(t)=e^t+e^{-t}\\y(t)=5-2t\end{cases}\implies\begin{cases}x'(t)=e^t-e^{-t}\\y'(t)=-2\end{cases}[/tex]
The length of the curve is given by the integral
[tex]\displaystyle\int_0^3\sqrt{(e^t-e^{-t})^2+(-2)^2}\,\mathrm dt[/tex]
Expand and rewrite the integrand:
[tex](e^t-e^{-t})^2+(-2)^2=e^{2t}+2+e^{-2t}[/tex]
[tex]=e^{-2t}(e^{4t}+2e^{2t}+1)[/tex]
[tex]=e^{-2t}(e^{2t}+1)^2[/tex]
[tex]\implies\sqrt{(e^t-e^{-t})^2+(-2)^2}=\dfrac{e^{2t}+1}{e^t}[/tex]
Now the integral is
[tex]\displaystyle\int_0^3\dfrac{e^{2t}+1}{e^t}\,\mathrm dt=\int_0^3(e^t+e^{-t})\,\mathrm dt[/tex]
[tex]=2\displaystyle\int_0^3\cosh t\,\mathrm dt[/tex]
[tex]=2\sinh t\bigg|_{t=0}^{t=3}[/tex]
[tex]=2\sinh 3[/tex]
