What is the antiderivative of 1/(x^2+4x+4) ?

To find the anti derivative of
1/(x^2+4x+4).

We know that x^2+4x+4 =
(x+2)^2.

Therefore Integral { 1/(x^2+4x+4) }dx = Integral
{1/(x+2)^2 } dx.

Now put (x+2) = t, then dx =
dt.

Therefore Integral {1/(x+2)^2} dx = Integral 1/t^2
dt

Integral {1/(x+2)^2} dx = Integral t^(-2)
dt

Integral {1/(x+2)^2} dx = (1/(-2+1)) t^(-2+1) +
Constant

Integral {1/(x+2)^2 } dx = (1/-1)(1/t)
+C

Integral {1/(1+x)^2}dx = -1/t
+C

Integral {1/(x+2)^2} = -1/(2+x) +C , as t =
2+x.

Therefore the antiderivative of 1/(x^2+4x+4) =Integral
{1/(x+2)^2} dx  = -1/(x+2) +C.