Answer
$$\eqalign{
& \int_0^e {\frac{{dx}}{{2x + e}}} \cr
& or \cr
& = \frac{1}{2}\int_0^e {\frac{{2dx}}{{2x + e}}} \cr
& {\text{integrate}} \cr
& \frac{1}{2}\int_0^e {\frac{{2dx}}{{2x + e}}} = \frac{1}{2}\left[ {\ln \left| {2x + e} \right|} \right]_0^e \cr
& {\text{evaluate}} \cr
& \frac{1}{2}\left[ {\ln \left| {2x + e} \right|} \right]_0^e = \frac{1}{2}\left[ {\ln \left| {2e + e} \right| - \ln \left| {2\left( 0 \right) + e} \right|} \right] \cr
& {\text{simplify}} \cr
& \frac{1}{2}\left[ {\ln \left| {2x + e} \right|} \right]_0^e = \frac{1}{2}\left[ {\ln \left| {3e} \right| - \ln \left| e \right|} \right] \cr
& \frac{1}{2}\left[ {\ln \left| {2x + e} \right|} \right]_0^e = \frac{1}{2}\left[ {\ln 3 + \ln e - \ln e} \right] \cr
& = \frac{1}{2}\ln 3 \cr} $$
Work Step by Step
$$\frac{1}{2}\ln 3$$
