If you mean
∫ 1/(t^2-1)^2 dt
try using partial fractions:
1/(t^2-1)^2 =
1/4 (1/(t+1) + 1/(t+1)^2 - 1/(t-1) + 1/(t-1)^2)
each of those is easy to integrate.
interge[1/(t*t-1)(t*t-1)]
I haven't got a way to solve it.. please help me..
1 answer