Complete the square for #x^2+2x+5#:
#x^2+2x+5=0#
#x^2+2x=-5#
#x^2+2x+(2/2)^2=-5+(2/2)^2#
(Recall that completing the square requires adding #(b/2)^2# to each side of our equation. Here, #b=2#.)
#x^2+2x+1=-5+1#
#(x+1)^2=-4#
#(x+1)^2+4=0# is our completed square.
Rewrite the integral using the completed square:
#int(x+3)/((x+1)^2+4)dx#
Split the integral up:
#intx/((x+1)^2+4)dx+int3/((x+1)^2+4)dx#
For both integrals, we can say #u=x+1# and therefore #du=dx# and #x=u-1# . Rewrite with the substitution:
#int(u-1)/(u^2+4)du+int3/(u^2+4)du#
We can once again split up this first integral:
#intu/(u^2+4)du-int(du)/(u^2+4)du+int3/(u^2+4)du#
We can now integrate all of these with basic formulas. Recall that
#intdx/(x^2+a^2)=1/atan^-1(x/a)+C#
and #intx/(x^2+a)dx=1/2ln|x+a|+C#
Thus, our integral becomes:
#1/2ln|u^2+4|-1/2tan^-1(u/2)+3/2tan^-1(u/2)+C#
#1/2ln(u^2+4)+tan^-1(u/2)+C#
(Note that we dropped the absolute value bars on our natural logarithmic function. This is because #u^4+4# is always #>0#; absolute value bars around it are redundant).
Write in terms of #x#:
#1/2ln((x+1)^2+4)+tan^-1((x+1)/2)+C#
#1/2ln(x^2+2x+5)+tan^-1((x+1)/2)+C#
