Complete the square:
#int sqrt(x^2+2x)dx = int sqrt(x^2+2x+1-1)dx = int sqrt((x+1)^2-1)dx#
Substitute #x+1 = sect#, #dx = sect tant dt#, with #t in (0,pi/2)#:
#int sqrt(x^2+2x)dx = sqrt(sec^2t-1) sect tant dt#
Using the trigonometric identity #sec^2t-1 = tan^2t# and as #tan t > 0# for #t in (0,pi/2)#:
#int sqrt(x^2+2x)dx = int sqrt(tan^2t) sect tant dt#:
#int sqrt(x^2+2x)dx = int sect tan^2t dt#
#int sqrt(x^2+2x)dx = int sect (sec^2-1)dt#
#int sqrt(x^2+2x)dx = int sec^3t dt - int sect dt #
Now the first integral is:
#int sect dt = int sect (sect+tant)/(sect+tant)dt#
#int sect dt = int (sec^2t+sect tant)/(sect+tant)dt#
#int sect dt = int (d(sect +tant))/(sect+tant)dt#
#int sect dt =ln abs (sect+tant) +c#
and for the second we can integrate by parts
#int sec^3t dt = int sect sec^2t dt = int sect d/dt(tant) dt#
#int sec^3t dt = sect tant - int tant d/dt (sect)dt#
#int sec^3t dt = sect tant - int sect tan^2t dt#
#int sec^3t dt = sect tant - int sect (sec^2-1)dt#
#int sec^3t dt = sect tant - int sec^3t dt + int sect dt#
The integral appears on both sides of the equation and we can solve for it:
#int sec^3t dt =(sect tant)/2 +1/2 int sect dt#
Putting it together:
#int sqrt(x^2+2x)dx = (sect tant)/2 - 1/2ln abs (sect+tant) +C#
To undo the substitution:
#sect = x+1#
#tant = sqrt(sec^2-1) = sqrt((x+1)^2 -1) = sqrt(x^2+2x)#
Finally:
#int sqrt(x^2+2x)dx = ((x+1)sqrt(x^2+2x))/2 -1/2ln abs((x+1) +sqrt(x^2+2x))+C#
