I faced some difficulties while trying to calculate $\int e^{-ax}$cos(bx)dx. I rewrote this as the integral of $e^{(-a + ib)x}$ calculating $\int e^{(-a + ib)x}dx$ (the answer is then the real part of the latter integral). I found $$\int e^{(-a + ib)x}dx = \frac{1}{-a+ib}e^{(-a + ib)x}.$$ My problem is that I don't know how to rewrite $\frac{1}{-a+ib}$, because when I try to use the basic rules for partial fractions, I get stuck. Thanks!