91Ó°ÊÓ

To evaluate the inner integral, we will integrate the function \(e^{x+y}\) with respect to \(x\) keeping \(y\) constant, and then evaluate the definite integral from \(x=0\) to \(x=\ln 3\). The integral with respect to x is: $$ \int_{0}^{\ln 3} e^{x+y} d x $$ To integrate, we have: $$ \int e^{x+y} dx = e^{x+y}*\frac{1}{y+1} $$ Now, we will evaluate this for x from 0 to \(\ln 3\) $$ \Big[ e^{x+y}*\frac{1}{y+1} \Big]_{0}^{\ln 3} = e^{\ln 3+y}*\frac{1}{y+1} - e^{0 + y}*\frac{1}{y+1} $$ This reduces to: $$ \frac{3e^y - e^y}{y+1} = \frac{2e^y}{y+1} $$ So, after evaluating the inner integral, we get: $$ \int_{1}^{\ln 5} \frac{2e^y}{y+1} dy $$

Now we will integrate the remaining integral with respect to \(y\) from \(y=1\) to \(y=\ln 5\). $$ \int_{1}^{\ln 5} \frac{2e^y}{y+1} dy $$ To integrate this expression, we can use substitution. Let \(u = y + 1\). So, \(du = dy\). Our integral becomes: $$ \int_{2}^{\ln 5 + 1} \frac{2e^{u-1}}{u} du $$ Now, we can integrate by multiplying through by \(2e^{-1}\) and integrating. $$ 2e^{-1}\int_{2}^{\ln 5 + 1} \frac{e^u}{u} du $$ This type of integral cannot be directly integrated in terms of elementary functions. However, we can express the result as an improper integral. \(\int \frac{e^u}{u} du\) is related to the exponential integral, a special function denoted by \(Ei(u)\). Therefore, our final answer is: $$ 2e^{-1}\left[ Ei(\ln 5 + 1) - Ei(2)\right] $$ Note that this is an exact representation of the given definite integral in terms of special functions.