Answer
$$\int e^{a x} \cos b x d x=\frac{e^{a x}(a \cos b x+b \sin b x)}{a^{2}+b^{2}}+C$$
Work Step by Step
Let
$$I= \int e^{a x} \cos b x d x$$
Use integration by parts , let
\begin{aligned}
u&= \cos b x \ \ \ \ \ \ &dv&= e^{a x} dx\\
du&=- b\sin b x dx \ \ \ \ \ \ &v&=\frac{1}{a} e^{a x}
\end{aligned}
then
\begin{aligned}
I &= uv-\int vdu \\
&= \frac{1}{a} e^{a x} \cos bx+\frac{b}{a}\int e^{a x}\sin bxdx \\
& = \frac{1}{a} e^{a x} \cos bx+\frac{b}{a}J\ \ \ \ \ \ \ (1)\\
\end{aligned}
Where
$$J=\int e^{a x}\sin bxdx$$
Use integration by parts , let
\begin{aligned}
u&= \sin b x \ \ \ \ \ \ &dv&= e^{a x} dx\\
du&= b\cos b x dx \ \ \ \ \ \ &v&=\frac{1}{a} e^{a x}
\end{aligned}
Then
\begin{aligned}
J&= uv-\int vdu\\
&= \frac{1}{a} e^{a x} \sin bx- \frac{b}{a}\int e^{ax} \cos bxdx\\
&= \frac{1}{a} e^{a x} \sin bx- \frac{b}{a}I\ \ \ \ \ \ \ \ \ \ \ (2)
\end{aligned}
It follows that from (1) and (2)
\begin{aligned}
I& \frac{1}{a} e^{a x} \cos bx+\frac{b}{a}J \\
& = \frac{1}{a} e^{a x} \cos bx+\frac{b}{a}\left[ \frac{1}{a} e^{a x} \sin bx- \frac{b}{a}I\right]\\
&= \frac{1}{a} e^{a x} \cos bx+\frac{b}{a^2}e^{a x} \sin bx-\frac{b}{a^2}I\\
\left(1+\frac{b^2}{a^2}\right)I&= e^{a x}[\frac{1}{a}\cos bx +\frac{b}{a^2}\sin bx]+C\\
\left( \frac{a^2+b^2}{a^2}\right)I&= e^{a x}[\frac{1}{a}\cos bx +\frac{b}{a^2}\sin bx]+C\\
\left( a^2+b^2 \right)I&= e^{a x}[a\cos bx +b\sin bx]+C\\
\end{aligned}
Hence
$$\int e^{a x} \cos b x d x=\frac{e^{a x}(a \cos b x+b \sin b x)}{a^{2}+b^{2}}+C$$
