Integral 3/t^2 dt?

Answer 1

Pull the 3 out of the integral because it's a constant scalar. So you get 3*INT[(t^(-2))dt], and you can use the power rule (I prefer "reverse power rule" since we're going the other way) to integrate t^(-2)dt.

Answer 2

int 3/t^2
=3*(-1)/t +c where c is any arbitrary constant
= -3/t + c

Answer 3

I = 3.∫ t^(-2).dt
I = 3. t^(-1) / (-1) + C
I = - 3 / t + C