Complete the square:
∫√x2+2xdx=∫√x2+2x+1−1dx=∫√(x+1)2−1dx
Substitute x+1=sect, dx=secttantdt, with t∈(0,π2):
∫√x2+2xdx=√sec2t−1secttantdt
Using the trigonometric identity sec2t−1=tan2t and as tant>0 for t∈(0,π2):
∫√x2+2xdx=∫√tan2tsecttantdt:
∫√x2+2xdx=∫secttan2tdt
∫√x2+2xdx=∫sect(sec2−1)dt
∫√x2+2xdx=∫sec3tdt−∫sectdt
Now the first integral is:
∫sectdt=∫sectsect+tantsect+tantdt
∫sectdt=∫sec2t+secttantsect+tantdt
∫sectdt=∫d(sect+tant)sect+tantdt
∫sectdt=ln|sect+tant|+c
and for the second we can integrate by parts
∫sec3tdt=∫sectsec2tdt=∫sectddt(tant)dt
∫sec3tdt=secttant−∫tantddt(sect)dt
∫sec3tdt=secttant−∫secttan2tdt
∫sec3tdt=secttant−∫sect(sec2−1)dt
∫sec3tdt=secttant−∫sec3tdt+∫sectdt
The integral appears on both sides of the equation and we can solve for it:
∫sec3tdt=secttant2+12∫sectdt
Putting it together:
∫√x2+2xdx=secttant2−12ln|sect+tant|+C
To undo the substitution:
sect=x+1
tant=√sec2−1=√(x+1)2−1=√x2+2x
Finally:
∫√x2+2xdx=(x+1)√x2+2x2−12ln∣∣(x+1)+√x2+2x∣∣+C