Answer
$\dfrac{3}{2}(5-e)$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{0}^{\ln 2}[e^{2x+y}]_{1}^{\ln 5} dx=\int_{0}^{\ln 2}[5e^{2x}-e^{2x+1}] dx$
or, $[\dfrac{5}{2}e^{2x}-\dfrac{1}{2}e^{2x+1}]_{0}^{\ln 2} =\dfrac{5}{2}e^{2 \ln 2}-\dfrac{1}{2}e^{2\ln 2+1}-\dfrac{5}{2}+\dfrac{1}{2}$
This implies that
$\dfrac{15}{2}-\dfrac{3}{2}e=\dfrac{3}{2}(5-e)$
