Using integrating factor I have the following differential equation

$$\frac{dN(t)}{dt} = A - \mu N(t)$$

I understand that I will need to use an integrating factor but am not sure how to proceed. I think I should use $e^{-\mu t}$ as the integrating factor but am not entirely sure.

I believe I should end up with 

$$N(t)=N_{0}e^{-\mu t}+\frac{A}{\mu}(1-e^{-\mu t})$$ where $N(0)=N_{0}$ is the initial condition.

Any help would be much appreciated!
You have $N'(t)+\mu N(t)=A$. Multiply by $e^{\mu t}$ to get $$(N(t)e^{\mu t})'=Ae^{\mu t}$$
Why not just separate the variables?

We have $$\frac{dN}{dt}=A-\mu N \iff\int\frac{1}{A-\mu N}dN=\int dt$$
$$\iff \int_{N_0}^{N}\frac{1}{A-\mu \nu}d\nu=\int_0^td\tau$$
$$\iff-\frac{1}{\mu}\bigg[\ln|A-\mu\nu|\bigg]_{N_0}^{N}=\bigg[\tau\bigg]_0^t$$
$$\iff-\frac{1}{\mu} \ln\left| \frac{A-\mu N}{A-\mu N_0}\right|=t \iff \frac{A-\mu N}{A-\mu N_0}=e^{-\mu t}$$
$$\iff \mu N=A-(A-\mu N_0)e^{-\mu t}$$

$$\iff \boxed{N(t)=\frac{A-(A-\mu N_0 ) e^{-\mu t}}{\mu}}$$