Answer
$$u\left( x \right) = {e^x} - 2{e^{ - 2x}} + \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& u'\left( x \right) = \frac{{{e^{2x}} + 4{e^{ - x}}}}{{{e^x}}} \cr
& u\left( x \right) = \int {u'\left( x \right)} dx \cr
& then \cr
& u\left( x \right) = \int {\left( {\frac{{{e^{2x}} + 4{e^{ - x}}}}{{{e^x}}}} \right)} dx \cr
& u\left( x \right) = \int {\left( {\frac{{{e^{2x}}}}{{{e^x}}} + \frac{{4{e^{ - x}}}}{{{e^x}}}} \right)} dx \cr
& u\left( x \right) = \int {\left( {{e^x} + 4{e^{ - 2x}}} \right)} dx \cr
& find{\text{ the general solution}} \cr
& u\left( x \right) = {e^x} - 2{e^{ - 2x}} + C \cr
& {\text{using the initial condition }}u\left( {\ln 2} \right) = 2 \cr
& 2 = {e^{\ln 2}} - 2{e^{ - 2\left( {\ln 2} \right)}} + C \cr
& 2 = 2 - 2\left( {\frac{1}{4}} \right) + C \cr
& C = \frac{1}{2} \cr
& {\text{the solution to the initial value problem is}} \cr
& u\left( x \right) = {e^x} - 2{e^{ - 2x}} + \frac{1}{2} \cr} $$
