Answer
$$\frac{1}{2}\tan^{-1}\frac{x-b}{2}+C$$
Work Step by Step
Given $$\int \frac{d x}{(x-b)^{2}+4}$$
Let $$u=x-b\ \ \ \ \ \ du=dx$$
Then
\begin{align*}
\int \frac{d x}{(x-b)^{2}+4}&=\int \frac{d u}{u^{2}+4}\\
&= \frac{1}{2}\tan^{-1}\frac{u}{2}+C\\
&= \frac{1}{2}\tan^{-1}\frac{x-b}{2}+C
\end{align*}
