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

### 2 Answers | Add Yours

We have to find the indefinite integral of y=1/(x^2+12x+36)

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

=> 1/(x + 6)^2

Int [( 1/(x + 6)^2) dx]

let x + 6 = y, dy = dx

=> Int [(1/y^2) dy]

=> y^(-2 + 1)/(-2 + 1)

=> -1/y

substitute y = x + 6

=> -1/(x + 6) + C

**The required integral is -1/(x + 6) + C**

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.**