Find the indefinite integral of y=1/(x^2+6x+9)?

We notice that the denominator is a 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 use the techinque of changing the
variable. For this reason we'll substitute x+3 by t.

x+3 =
t

We'll differentiate both sides with respect to
x:

(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) = t^(-1)/-1 =
-1/t

But t =
x+3

The indefinite integral of the given
function is represented by the primitive F(x) = -1/(x+3) +
C.