Integrating, Long Division, Completing the Square Practice Test

$\frac{1}{x^2+4x+8}+C$

$\ln(x^2+4x+8)+C$

$\arctan!\left(\frac{x+2}{2}\right)+C$

$\frac{1}{2}\arctan!\left(\frac{x+2}{2}\right)+C$

$\frac{1}{2}\ln!\left|\frac{x+2}{2}\right|+C$