# How do you integrate [(x + 4)/(x^2 + 8 x +17)] dx?

$\frac{x + 4}{{x}^{2} + 8 x + 17} = \frac{x + 4}{{x}^{2} + 8 x + 16 + 1} = \frac{x + 4}{{\left(x + 4\right)}^{2} + 1}$
Now with $u = x + 4$ this becomes $\frac{u}{{u}^{2} + 1}$ whose integral (by another substitution) is
$\frac{1}{2} \ln \left({u}^{2} + 1\right) + C$
Finish by going back to the variable $x$.