# What is the antiderivative of 1/(x^2+4x+4) ?

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### 2 Answers

We notice that the denominator is 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 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 anti derivative of 1/(x^2+4x+4).

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

Therefore Integral { 1/(x^2+4x+4) }dx = Integral {1/(x+2)^2 } dx.

Now put (x+2) = t, then dx = dt.

Therefore Integral {1/(x+2)^2} dx = Integral 1/t^2 dt

Integral {1/(x+2)^2} dx = Integral t^(-2) dt

Integral {1/(x+2)^2} dx = (1/(-2+1)) t^(-2+1) + Constant

Integral {1/(x+2)^2 } dx = (1/-1)(1/t) +C

Integral {1/(1+x)^2}dx = -1/t +C

Integral {1/(x+2)^2} = -1/(2+x) +C , as t = 2+x.

Therefore the antiderivative of 1/(x^2+4x+4) =Integral {1/(x+2)^2} dx = -1/(x+2) +C.