There is no way to really perform partial fraction decomposition here; however, we may integrate using the following substitution:
#u=x+1#
#x=u-1#
#du=dx#
#int(u-1+4)/(u^2+4)dx=int(u+3)/(u^2+4)du#
We may split this up as follows:
#intu/(u^2+4)du+3int(du)/(u^2+4)#
#intu/(u^2+4)du=1/2ln(u^2+4)# -- this can be solved with a mental substitution, as the differential of #u^2+4# shows up in the numerator with a bit of simplification. We do not attach absolute value bars as #u^2+4# is always positive.
#3int(du)/(u^2+4)=3/2arctan(u/2)#.
This comes from the common integral, #intdx/(x^2+a^2)=1/aarctan(x/a)#. For our case, #a=sqrt4=2#.
Thus, rewriting in terms of #x# and simplifying the logarithm, we see
#int(x+4)/((x+1)^2+4)dx=lnsqrt[(x+1)^2+4]+3/2arctan((x+1)/2)+C#
