# Exponential integration rule

Given the function $e^ae^{-ba}$

The indefinite integral is $\int e^ae^{-ba} \mathrm da = \int e^{a-ba} \mathrm da$ is $\frac{e^ae^{-ba}}{1-b}$

I get that $\int e^u = e^u\frac{\mathrm du}{\mathrm da}$.

I cannot seem to understand how the $(1-b)$ term ends up in the denominator.

Can someone point out the rule that I am missing that puts the $(1-b)$ in the denominator?

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Briefly, it comes from $e^a e^{-ba} = e^{a-ba} = e^{(1-b)a}$. – johnny Nov 23 '11 at 17:41
Differentiate $e^{a-ba}$. How does it compare to your integrand? – David Mitra Nov 23 '11 at 17:43

Note that $a-ba=(1-b)a$. So $$\int e^ae^{-ba}\,da = \int e^{a-ba}\,da = \int e^{(1-b)a}\,da.$$ Setting $u=(1-b)a$, we have $du = (1-b)da$, or $\frac{1}{1-b}\,du = da$. Hence $$\int e^{a}e^{-ba}\,da = \int e^{(1-b)a}\,da = \int \frac{1}{1-b}e^u\,du = \frac{1}{1-b}e^u+C = \frac{e^{(1-b)a}}{1-b}+C.$$
I don't understand what you mean by "$\int e^u = e^u\frac{du}{da}$" (there seem to be lots of things missing, like a $d\Box$ in the integral and a constant of integration).
There are probably things missing which is probably aiding in my confusion. What I was missing is that $du=(1-b)da; \frac{1}{1-b}du=da$ Thank you. – strimp099 Nov 23 '11 at 17:59