dy/dx = 1/(x^2 + 6x + 9)

We need to find y.

Then we will need to integrate dy/dx

==> dy = 1/(x^2 + 6x+9) dx

We know that x^2+ 6x+9 = (x+3)^2

==> dy = 1/(x+3)^2 dx

Now we will find the integral.

==> Int dy = Int 1/(x+3)^2 dx

==> Let u = x+3 ==> du = dx

==> y= Int 1/u^2 du = Int u^-2 du = u^-1/-1 = -1/u

==> y= -1/u + C

==> y = -1/(x+3) + C

**Then the anti-derivative is y= -1/(x+3) + C**

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.**