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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:

$<br /> <br /> \int \frac 1 {(2t+1)^2} dt<br /> <br />$

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.