`int(2 dx)/(x^2+8x+17`

Now

`x^2+8x+17`

`=(x^2+8x+16)+1`

`=(x+4)^2+1`

So if we let `u = x + 4`, then `du = dx` and we have:

`int(2\ dx)/(x^2+8x+17)`

`=2int(du)/(u^2+1)`

`=2\ tan^-1u+K`

`=2\ tan^-1(x+4)+K`

Of course, we could also write this answer as

`2\ arctan(x+4)+K`