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

Answer 1

∫ ( -e^(1/t) / t² ) dt

This is actually really quite easy. Keeping the reverse chain rule in mind, if you have the following, integrals end up very easy:

∫ [ f'(g(t)) · g'(t) ] dt = f(g(t)) + C

If you go ahead and define the following functions, you have exactly that form:

f(t) = e^(t)
g(t) = 1/t

Therefore, f'(t) = e^t, and g'(t) = -1/t², so we have exactly the form above. So, your answer is simply:

∫ ( -e^(1/t) / t² ) dt = e^(1/t) + C.

Answer 2

one technique is to look at the terms you're trying to integrate, and make a substitution
u=f(t) such that du = f'(t) dt where f'(t) cancels off terms in the original integral.

In your case, if you use u = 1/t,
du/dt will have a 1/t^2 term which will take out your t^2 in the denominator in the original integral. Try this, and after adjusting for the constant, see if you can solve it.

Answer 3

Use substitution of variables:

u= 1/t ==> du = -1/(t^2)

then int((-e^(1/t))/(t^2)dt = int(e^udu) = e^u + C

Replacing back the variable: = e^(1/t) + C

Answer 4

Doesn't look all that hard. Let 1/t= u and
du= -1/t^2 * dt. Doesn't that sort of look like a
e^u * du form?

Answer 5

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

Answer 6

way over my head buddy