# Evaluate the indefinite integral

• October 15th 2010, 08:21 PM
auntjamima
Evaluate 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
• October 15th 2010, 08:35 PM
Educated
$\displaystyle \int \dfrac{dt}{(t^2 - 1)^2} = \int \dfrac{dt}{(1-t^2)^2} =$

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

EDIT: Actually, Nevermind.
• October 15th 2010, 08:41 PM
harish21
Use partial fractions first and then integrate:

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