What is the indefinite integral of the function y=1/(x^2+12x+36) ?

Since the denominator is the perfect square, we can
re-write the given function:

1/(x^2+12x+36) =
1/(x+6)^2

We'll re-write the
integral:

Int f(x)dx = Int
dx/(x+6)^2

We'll use substitution to solve the
integral:

x+6 = t

We'll
differentiate both sides:

(x+6)'dx = dt => dx =
dt

We'll re-write the integral in
t:

Int dx/(x+6)^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+6

The requested indefinite integral of the
function is Int dx/(x^2+12x+36) = -1/(x+6) + C.