Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 29

Integrate int(x+4)/(x^2+2x+5)dx
int(x+4)/(x^2+2x+5)dx=int(x+1)/(x^2+2x+5)dx+int3/(x^2+2x+5)dx

Integrate the first integral on the left side of the equation using the u-substitution method.
Let u=x^2+2x+5
(du)/(dx)=2x+2
dx=(du)/(2(x+1))
int(x+1)/(x^2+2x+5)dx
=int(x+1)/u*(du)/(2(x+1))
=1/2ln|u|+C
1/2ln|x^2+2x+5|+C

The second integral on the left side will match the form
intdx/(x^2+a^2)=1/atan^-1(x/a)+C after you complete the square in the denominator.

int3/(x^2+2x+5)dx
=3intdx/[(x^2+2x+1)+5-1]
=3intdx/[(x+1)^2+2^2]
=3(1/2)tan^-1((x+1)/2)+C
=(3/2)tan^-1[(x+1)/2]+C

The final answer is:
1/2ln|x^2+2x+5|+(3/2)tan^-1[(x+1)/2]+C