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