View Full Version : Integration, either I'm losing it or I've been working too long
albear
Jun 17, 2007, 08:27 AM
I need to integrate 1000(2t+1)^-2
I thought it would go [(1000(2t+1)^-1)/-1 ] * t
galactus
Jun 17, 2007, 08:56 AM
Let u=2t+1, \;\ du=2dt, \;\ \frac{du}{2}=dt
This gives:
500\int\frac{1}{u^{2}}du
albear
Jun 17, 2007, 09:15 AM
sorry how do you apply dt=du/2 to get 500(integral)(1/u^2)du
albear
Jun 17, 2007, 11:16 AM
Hello
ebaines
Jun 18, 2007, 01:39 PM
You are trying to solve:
\int \frac 1 {(2t+1)^2} dt
correct?
So replacing 2t+1 by u, and dt by du/2 gives you the form that Gallactus suggests. Now you should be able to integrate that form, and then the last step is to substitute 2t+1 for u to get the final answer.