Substitute #w=tan^-1(x+1)#, #(dw)/dx=1/(x^2+2x+2)#.
The integral becomes #int cos^2w dw#
#=int(1+cos 2w)/2 dw#
#=w/2+(1/2)sin w cos w#
#=(1/2)tan^-1(x+1)+(1/2).(x+1)/sqrt(x^2+2x+2)xx1/sqrt(x^2+2x+2)+C#
#=(1/2)tan^-1(x+1)+(1/2)(x+1)/(x^2+2x+2)+C#
It is helpful to draw a right triangle with angle #tan^-1(x+1)#, adjacent side #1#, opposite side #x+1# and hypotenuse #x^2+2x+2#. And remember that #cos 2w = 2cos^2w-1#, #sin 2w = 2sin w cos w#.
If the initial substitution is hard to spot, substitute #u=x+1# first in order to get rid of the #2x#, giving #int (du)/(u^2+1)^2# then further substitute #u=tan w#, and draw a triangle with sides #1#, #u#, #sqrt(1+u^2)#
