Evaluate the following integral: `int_1^2 1/(x^2)e^(-1//x)dx`

Explanation

Given:`int_1^2 1/(x^2)e^(-1//x)dx`

`I=int_1^2 1/(x^2)e^(-1//x)dx`

Let `u=-1/x implies du=1/(x^2)dx`

When `x=1, u=−1` and when `x=2,u=− 1/2`

`I=int_(-1)^(1/2) e^udu`

`I=-int_(-1/2)^(-1) e^udu`

`I=-[e^u]_(-1/2)^(-1)`

`I=-[e^(-1)-e^(-1/2)]`

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