Integral of -e^(-x)

You have been given:

∫eudu=eu+C

where C is an undetermined constant of integration.
You have come upon the integral:

∫f(g(x))dx

where you know the indefinite integral of f(u), but not f(g(x)). In your case f(u) = e^u and g(x) = -x. If we go back to derivatives, we may remember a related rule called the chain rule:
(f(g(x))' = f'(g(x))*g'(x)
In other words:

∫f′(g(x))∗g′(x)dx=f(g(x))+C

In some texts, they will use shorthand Leibnitz notation: Let u = g(x). Then du = g'(x) dx. The above integral is then written:

∫f′(u)du=f(u)+C

This is usually called u-substitution. Note that you have to have both f'(g(x)) = f'(u) and du = g'(x) dx in the integrand. Sometimes you will have to multiply the integrand by a creative version of 1 in order to make this happen. In your example, let f'(u) = e^u since we already know how to integrate that and of course u = -x. Then du = -1 dx. Then we need our integrand to be f'(u) du = -e^-x dx. Unfortunately, our integrand is actually -(e^^-x dx) = -f'(u) du. Luckily, the -1 can be factored out of the integrand as it is a constant, so we have 

−∫f′(u)du=−f(u)+C=−eu+C