posted by W .
How do I integrate (e^(3/x))/(x^2) from 1 to 3. I understand how and why I need to do u-substitution, with u=3/x and so dx=(-3 du)/x^2. After plugging (-3 du)/x^2 in for dx, I get a really messy problem that I don't know how to finish. Can you please explain the step?
Thanks for your help!
I think you may be going about things backwards. If
u = 3/x
du = -3/x^2 dx
and your integrand becomes e^u du/(-3) = -1/3 Int(e^u du) = -1/3 e^u = -1/3 e^3/x
check: taking the derivative of -1/3 e^3/x we get
-1/3 * e^3/x * -3/x^2 = e^(3/x) / x^2
So, evaluating at 3 and 1,
-1/3 e^(3/3) + 1/3 e^(3/1) = 1/3 (e^3 - e)