Wha is the integral of 1/t^2 ?

Answer 1

Well the integral of 1/t^2, is can be written as the integral of t^-2, now for any integral of a polynomial
(x^n), n cannot equal -1, cause the integral of that is ln(x), but the integral of (x^n), is [x^(n+1)]/[n+1], in your case n=-2, so it is just [t^(-2+1)]/[-2+1], which is just -t^-1, so the integral of 1/t^2, is just
(-t^-1)+C

Answer 2

integral 1/t^2 dt= integral [1 * t^-2] dt = integral [t^-2] dt
n=-2

with integral x^n dx = (1/n+1) * x^(n+1) +C


integral [t^-2] dt =(1/-2+1)* t^(-2+1) +C

= -1 * (t^-1) +C

= (-1/t) +C

Answer 3

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