Evaluate the improper integral or state that it diverges: integral from 6 to infinity (1/t^2-5t)dt. I need help on solving this and what does it mean by converges and diverges?

Answer 1

integral = -(2t-5)/(t^2-5)^2

at infinity ---> -2t/t^2 = 2/t = 0
at 6 = (5-2)/31^2
= .00312

It does not diverge but has a finite answer. If t = sqrt(5) were included, we would have had a problem with a zero denominator.

Answer 2

I think you missed a t in the denominator

∫ 1/(t^2-5t) dt = 1/5 log((5-t)/t)
the definite integral is log(6)/5

Answer 3

Whoops, sorry, use Steve' result.