Find the indefinite integral of y=1/(x^2+6x+9)?
We have to find the indefinite integral of y=1/(x^2+6x+9)
=> y = 1/(x + 3)^2 = (x + 3)^-2
Int[ y dx] = Int [(x + 3)^-2 dx]
let x + 3 = z, dz = dx
=> Int[z^-2 dz]
substitute z = x + 3
=> -1/(x + 3) + C
The required integral of y=1/(x^2+6x+9) is -1/(x + 3) + C
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.