# What is the indefinite integral of the function f(x)=1/(x^2+8x+16)?

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### 1 Answer

We notice that the denominator is a perfect square: x^2 + 8x + 16 = (x+4)^2

We'll re-write the integral:

Int f(x)dx = Int dx/(x+4)^2

We'll apply the techinque of substitution of the variable.

We'll replace x+4 by t.

x+4 = t

We'll differentiate both sides:

(x+4)'dx = dt => dx = dt

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

Int dx/(x+4)^2 = Int dt/t^2

Int dt/t^2 = Int [t^(-2)]*dt

Int [t^(-2)]*dt = t^(-2+1)/(-2+1) + C

Int [t^(-2)]*dt = t^(-1)/-1 + C

Int [t^(-2)]*dt = -1/t + C

But t = x+4

**The requested indefinite integral of the function is: Int dx/(x^2 + 8x + 16) = -1/(x+4) + C**** **