Originally Posted by
HallsofIvy You surely do not mean "$\displaystyle u= e^2$" then? That is a constant. If you mean "$\displaystyle u= e^{2x}$" then $\displaystyle du= 2e^{2x}dx$. But you don't have "$\displaystyle e^{2x}$ in the numerator to use with the "dx".
Instead, write the integral as $\displaystyle \int_0^{ln(3)}\frac{e^x}{e^{2x}}dx- \int_0^{ln(3)}\frac{1}{e^{2x}}dx= \int_0^{ln(3)} e^{-x}dx- \int_0^{ln(3)} e^{-2x}dx$ and do the two integrals separately. If you need to, let u= -x in the first integral and let u= -2x in the second integral.