How do you integrate #int (x+4)/(x^2+2x+4)# using trigonometric substitution?

We have:

#int(x+4)/(x^2+2x+4)dx=int(x+4)/((x+1)^2+3)dx#

Let #x+1=sqrt3tantheta#. Thus, #dx=sqrt3sec^2thetad theta#. Also note that this means #x+4=3+sqrt3tantheta#. This yields:

#=int(3+sqrt3tantheta)/(3tan^2theta+3)(sqrt3sec^2thetad theta)#

#=int(3sqrt3+3tantheta)/(3(tan^2theta+1))(sec^2thetad theta)#

Cancel the #3# terms and recall that tan^2theta+1=sec^2theta#:

#=int(sqrt3+tantheta)/sec^2theta(sec^2thetad theta)#

#=int(sqrt3+tantheta)d theta#

These are common integrals:

#=sqrt3theta+lnabssectheta+C#

Note that #theta=arctan((x+1)/sqrt3)#. Also, we can draw the right triangle where #tantheta=(x+1)/sqrt3# to see that #sectheta=sqrt((x+1)^2+3)/sqrt3=sqrt((x^2+2x+4)/3)#. Thus:

#=sqrt3arctan((x+1)/sqrt3)+lnabssqrt((x^2+2x+4)/3)+C#

Using the log rule #log(A^B)=Blog(A)#:

#=sqrt3arctan((x+1)/sqrt3)+1/2lnabs((x^2+2x+4)/3)+C#

Using the #log(A/B)=log(A)-log(B)# the #-1/2ln(3)# will come out of the natural logarithm and be absorbed into #C#:

#=sqrt3arctan((x+1)/sqrt3)+1/2lnabs(x^2+2x+4)+C#

Since #x^2+2x+4>0# for all (real) values of #x#:

#=sqrt3arctan((x+1)/sqrt3)+1/2ln(x^2+2x+4)+C#