I need to evaluate this integral

∫dt/(t^2-1)^2

I originally tried trig sub, by setting t=secx, but I only got so far as secx/(tanx)^3.

Any help would be greatly appreciated.

Thanks,

Mike

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- Oct 15th 2010, 08:21 PMauntjamimaEvaluate the indefinite integral
I need to evaluate this integral

**∫dt/(t^2-1)^2**

I originally tried trig sub, by setting t=secx, but I only got so far as secx/(tanx)^3.

Any help would be greatly appreciated.

Thanks,

Mike - Oct 15th 2010, 08:35 PMEducated
$\displaystyle \displaystyle \int \dfrac{dt}{(t^2 - 1)^2} = \int \dfrac{dt}{(1-t^2)^2} =$

Substitute in $\displaystyle t^2 = \sin^2(x)$ and use the pythagorean identity.

EDIT: Actually, Nevermind. - Oct 15th 2010, 08:41 PMharish21
Use partial fractions first and then integrate:

$\displaystyle \displaystyle{\frac{1}{(t^{2}-1)^2} = \frac{1}{4(t+1)}- \frac{1}{4(t-1)}+ \frac{1}{4(t+1)^2}+ \frac{1}{4(t-1)^2}}$