Answer
$16\sinh{2} -16\sinh{1}=8[e^2-e^{-2}-e+e^{-1}]$
Work Step by Step
Given: $\int^{4}_{1} \dfrac{ 8\cosh (\sqrt x)}{\sqrt x} dx$
Re-write as: $\int^{4}_{1}(16\cosh (\sqrt x))( \dfrac{ dx}{ 2\sqrt x} )$
Plug in: $\sqrt x= t$ and $dt=\dfrac{ dx}{ 2\sqrt x} $
Thus, the limits of integration will also get changed.
Then, $\int^{4}_{1}(16\cosh (\sqrt x))( \dfrac{ dx}{ 2\sqrt x}) =\int^{2}_{1} 16\cosh t dt$
or, $=[16 \sinh t]^{2}_{1}$
or, $= 16\sinh ( 2) -16\sinh (1)$
$=16[(\frac{e^2-e^{-2}}{2})-(\frac{e-e^{-1}}{2})]$
$=8[(e^2-e^{-2})-(e-e^{-1})]$
$=8[e^2-e^{-2}-e+e^{-1}]$
