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.