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

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the indefinite integral of y=1/(x^2+6x+9)

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]

=> z^-1/-1

=> -1/z

substitute z = x + 3

=> -1/(x + 3) + C

The required integral of y=1/(x^2+6x+9) is -1/(x + 3) + C

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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