Answer
$\displaystyle \frac{1}{2}\ln\left(x^{2}+2x+5\right)+\frac{3}{2}\tan^{-1}\left(\frac{x+1}{2}\right)+C$
Work Step by Step
$x^{2}+2x+5$ can not be factored ($b^{2}-4ac$ is negative), so we can't use the partial fraction method.
We transform the numerator:
$x+4=\displaystyle \frac{1}{2}(2x+2)+3$
$\displaystyle \frac{x+4}{x^{2}+2x+5}=\frac{1}{2}\cdot\frac{2x+2}{x^{2}+2x+5}+\frac{3}{x^{2}+2x+5}$
The second numerator is transformed by completing the square
$x^{2}+2x+5=x^{2}+2x+1+4=(x+1)^{2}+2^{2}$
$\displaystyle \int\frac{x+4}{x^{2}+2x+5}dx=\frac{1}{2}\int\frac{2x+2}{x^{2}+2x+5}dx+3\int\frac{dx}{(x+1)^{2}+2^{2}}$
$...\left[\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\tan^{-1}(\frac{u}{a})\right]$
$=\displaystyle \frac{1}{2}\ln\left(x^{2}+2x+5\right)+\frac{3}{2}\tan^{-1}\left(\frac{x+1}{2}\right)+C$
