Integral of dy/e^y dx

1. Let $u = e^{y}$.

   Then let $du = e^{y} dy$ and substitute $du$:

   $$\int \frac{1}{u^{2}}\, du$$

   1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

   Now substitute $u$ back in:

   $$- e^{- y}$$

2. Add the constant of integration:

   $$- e^{- y}+ \mathrm{constant}$$

The answer is:

$$- e^{- y}+ \mathrm{constant}$$