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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$\displaystyle \displaystyle \int \dfrac{dt}{(t^2 - 1)^2} = \int \dfrac{dt}{(1-t^2)^2} =$ Substitute in $\displaystyle t^2 = \sin^2(x)$ and use the pythagorean identity. EDIT: Actually, Nevermind.
Use partial fractions first and then integrate: $\displaystyle \displaystyle{\frac{1}{(t^{2}-1)^2} = \frac{1}{4(t+1)}- \frac{1}{4(t-1)}+ \frac{1}{4(t+1)^2}+ \frac{1}{4(t-1)^2}}$
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