Substitution problem with $\int e^{-1 \over x}{dx\over x²}$

Question

I am trying to comprehend what is happening in this integration problem. I have tried substituting for both $U = {-1 \over x}$ and $U = x²$ but I am not ending up with the correct result on both cases. This is the worked example:

$$\int e^{-1 \over x}{dx\over x²} = \int e^{-1 \over x}d{-1 \over x} = e^{-1 \over x} +C$$

Can anyone help me finding the correct substitution method? Thanks in advance!

Answer 1

The correct substitution is $u=-\frac{1}{x}$ given that $du=\frac{1}{x^2}$, so the integral is just $\int e^u du=e^u+C$.

Answer 2

Substitute $u=-\frac1x,\ du=\frac1{x^2}dx$

$$\int \frac{e^{-1/x}}{x^2}dx=\int e^{-1/x}\cdot\frac{1}{x^2}dx=\int e^u\,du=e^u+C=e^{-1/x}+C$$

Answer 3

If you do $$ u=-\frac{1}{x} $$ you also have $$ x=-\frac{1}{u},\qquad dx=\frac{1}{u^2}\,du $$ so the integral becomes $$ \int e^u\cdot u^2\cdot\frac{1}{u^2}\,du=\int e^u\,du $$