The antiderivative of a function is equivalent of the integral of that function.

Therefore the antiderivative of `(x^2+2x+2)^(1/2)` with respect to `x` is equivalent to

`int (x^2+2x+2)^(1/2)dx`

Rewriting `x^2 + 2x + 2` as `(x+1)^2 + 1` this is equal to

`int [(x+1)^2 + 1]^(1/2)dx`

Now make the substitution

`tan(t) = x+1` , where `(dx)/(dt) = sec^2t` . The integral is then equal to

`int (tan^2t + 1)^(1/2)sec^2t dt`

Since `tan^2t +1 = sec^2t` (trigonometric identity) this equals

`int sec^3t dt `

`= 1/2 sec(t)tan(t) + 1/2ln|sec(t) +tan(t)| + c`

`= 1/2sqrt(x^2+2x+2)(x+1) + 1/2ln|sqrt(x^2+2x+2) + (x+1)| + c`

(since `sec(arctan(x+1)) = sqrt((x+1)^2+1) = sqrt(x^2+2x+2)` )

This result is found by using integration by parts where we let

`u = sec(x)` and `v = tan(x)` and solve

`int u (dv)/(dx) = uv - int v (du)/(dx)`

**The antiderivative of (x^2+2x+2)^(1/2) is equivalent to the integral of the same expression. ****Using substitution and then integration by parts this is found to be **

**1/2(x^2+2x+2)^(1/2)(x+1) + 1/2 ln|(x^2+2x+2)^(1/2) + (x+1)| + c**