Evaluate the antiderivative of s(t)=1/(t^2+4t+4)?

2 Answers

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the anti derivative of s(t)=1/(t^2+4t+4).


s(t) = 1/(t^2+4t+4).

=> s(t) = 1/(t+2)^2.

Therefore Int s(t) dt =  Int {1/(t+2)^2} dt +C= Int (t+2)^(-2) dt +C= {(t+2)^(-2+1)}/(-2+1) +C= -(t+2)^(-1)+C.

Therefore Int s(t) dt = -1/(t+2) +C, where C is a constant of integration.

For s(t) = 1/(t^2+4t+4), the antiderivative = -1/(t+2)+C.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To evaluate the antiderivative, we'll have to determine the indefinite integral of the function.

We notice that the denominator is a perfect square:

t^2 + 4t + 4 = (t+2)^2

We'll re-write the integral:

Int s(t)dt = Int dt/(t+2)^2

We'll use the techinque of changing the variable.

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

t+2 = u

We'll differentiate both sides:

(t+2)'dt = du

So, dt = du

We'll re-write the integral in the variable u:

Int dt/(t+2)^2 = Int du/u^2

Int du/u^2 = Int [u^(-2)]*du

Int [u^(-2)]*du = u^(-2+1)/(-2+1) + C = u^(-1)/-1 + C = -1/u + C

But u = t+2

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