What is the antiderivative y of dy/dx=1/(x^2+6x+9) ?

We notice that the denominator is the perfect square:
x^2 + 6x + 9 = (x+3)^2

We'll re-write the
integral:

Int f(x)dx = Int
dx/(x+3)^2

We'll replace x+3 by
t.

x+3 = t

We'll differentiate
both sides:

(x+3)'dx = dt

So,
dx = dt

We'll re-write the integral in
t:

Int dx/(x+3)^2 = Int
dt/t^2

Int dt/t^2 = Int
[t^(-2)]*dt

Int [t^(-2)]*dt = t^(-2+1)/(-2+1) + C =
t^(-1)/-1 + C = -1/t + C

But t =
x+3

The requested antiderivative of dy/dx is
y = Int dx/(x+3)^2 = -1/(x+3) + C.