Need to understand logarithmic integrals

Question

So $d/dx \ln(x) = 1/x$. Why is $\int \frac 1 {e^y}\,dy \ne \ln(e^y)$?

Answer 1

Well

$$\int {dx\over x} = \log |x|+C$$

yes, but if you mean to integrate

$$\int {dy\over e^y}$$

in the second one the reason is obvious: it's not ${d(e^y)\over e^y}$ in the integrand. In that case you get:

$$\int {d(e^y)\over e^y} = \int {e^y\over e^y}\,dy = y + C = \log|e^y|+C$$

as desired.