Answer
$8(e^{2}-e^{-2}-e+e^{-1})$
Work Step by Step
Evaluate the integral as follows:
$\displaystyle \int_{1}^{4}\frac{8\cosh\sqrt{x}}{\sqrt{x}}dx$= ... $\qquad\left[\begin{array}{ll}
t=\sqrt{x} & dt=\frac{dx}{2\sqrt{x}}\\
x=1 & \rightarrow t=1\\
x=4 & \rightarrow t=2
\end{array}\right]$
$=\displaystyle \int_{1}^{2}8\cosh t(2dt)=16\int_{1}^{2}\cosh tdt=\qquad$
Using table 8:
$=16[\sinh t]_{1}^{2}$
$=16(\sinh 2-\sinh 1)$
$=16[(\displaystyle \frac{e^{2}-e^{-2}}{2})-(\frac{e-e^{-1}}{2})]$
$=8(e^{2}-e^{-2}-e+e^{-1})$
