what @robjohn said is true:
because of the nature of this equation it is separable so an integrating factor is not strictly necessary. but because you requested it be done in that method:
first consider the form of the equation for an integrating factor:
A differential equation of the form $y' + P(x)y = Q(x) $ can be solved by
$ y = \frac{\int{\mu Q(x)} dx}{\mu} + \frac{k}{\mu} $ where $k$ is an arbitrary constant and $\mu $ is the integrating factor ($\mu = e^{\int{P(x)}dx}$)
Let's fit your equation into the form needed for the method of integrating factors (In your case, $y$ is $c$ ):
$ c' + c(1+a) = 1 $
so you should obtain for an integrating factor of $ \mu = e^{\int{(1+a)}dt} = e^{t(1+a)} $
so $ c = \frac{1}{e^{t(1+a)}} \int{(e^{t(1+a)})dt} + \frac{k}{e^{t(1+a)}}$
if you integrated correctly, you should get $\int{(e^{t(1+a)})dt} = \frac{1}{1+a}e^{t(1+a)}$ which you will find nicely cancels with the integrating factor, $\mu$ which is already in the denominator. So your final answer should be
$ c = \frac{1}{1+a} + \frac{k}{e^{t(1+a)}} = \frac{1}{1+a} + ke^{-t(1+a)} $