# Find the indefinite integral of y=1/(x^2+4x+4)?

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### 3 Answers

We have to find the integral of y=1/(x^2+4x+4).

Now we see that y = 1 / (x^2+4x+4)

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

Let us use the notation p = x+2

We see that dp = dx

=> Int [y/dy ]= Int [dx/ (x+2)^2]

=> Int [ dp / p^2]

Now the integral of x^n = x^(n+1)/ (n+1)

=> Int [ dp / p^2]

=> p^-1/-1

=> -1/p

But p = x+2

**Therefore the required integral is -1/(x+2).**

We'll write the denominator as the result of expanding the square: x^2 + 4x + 4 = (x+2)^2

We'll re-write the integral:

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

We'll use the techinque of changing the variable.

For this reason we'll substitute x+2 by t.

x+2 = t

We'll differentiate both sides:

(x+2)'dx = dt

So, dx = dt

We'll re-write the integral in the variable t:

Int dx/(x+2)^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+2

**Int dx/(x+2)^2 = -1/(x+2) + C**** **

To find the integral of y = 1/x^2+4x+4).

We know that 1/(x^2+4x+4) = 1/(x+2)^2.

Therfore Integral y dx = Integral {dx/(x^2+4x+4)}

Integral y dx = Integral dx/(x+2)^2.

Integral y dx = Integral (x+2)^(-2) dx...(1)

Let us have a transformation (x+2) = t

Then differentiating both sides, we get:

dx = dt..

Therfore Integral (x+2)^(-2) dx = integral t^(-2) dt.

Integral (x+2)^(-2) dx = (1/(-2+1))t ^(-2+1) +Const.

Intagral (x+2)^(-2) dx = - t^(-1).

Integral (x+2)^(-2) = - 1/(x+2) +C .

Therefore from (1) ,

Integral ydx = Integral dx/(x+2)^2 = -1/(x+2) + C.