Firstly, note that


If we put `u = x + 1`, then `du = dx` and our integral becomes:


Now, we use `u = sec\ θ` and so `du = sec\ θ\ tan\ θ\ dθ`

The square root becomes:

`sqrt(u^2-1)` `=sqrt(sec^2theta-1)` `=sqrt(tan^2 theta)` `=tan\ theta`

Returning to our integral, we have:


`=int(sec\ theta\ tan\ theta\ d theta)/(tan\ theta)`

`=int sec\ theta\ d theta`

`=ln\ |sec\ theta+tan\ theta|+K`

`=ln\ |x+1+sqrt(x^2+2x)|+K`