What is the indefinite integral of ( e ^ (1 / t) ) / ( t ^ 2 ) dt ?

Answer 1

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Answer 2

Since the derivative of 1/t is -t^(-2), we need that term in the integral so that we can fully integrate the e^(1/t). Think of it as the derivative in reverse, but usign the chain rule along the way.

∫ ( e^(1/t) ) / (t^2) dt =
∫ (t^-2) e^(1/t) dt =
-∫ -(t^-2) e^(1/t) dt =
- e^(1/t) + c

You could also use "u-substitution" by letting u = 1/t, so that du = -t^(-2) dt, so dt = du / -t^(-2). Pluging these back in gives
∫ e^u / (-t^(2) t^2 ) du = -∫ e^u du = -e^u + c = -e^(1/t) + c

Answer 3

Let u = 1 / t
du = - t ^(-2) dt = - (1 / t² ).dt
I = - ∫ e^(u) du
I = - e^u + C
I = - e^(1/t) + C