Question

How do you integrate `int e^(1/x^2)/x^3dx`?

Answer 1

`-e^(1//x^2)/2+C`

Explanation:

`inte^(1//x^2)/x^3dx`

Let `u=1/x^2`. Differentiating this, we see that `du=-2/x^3dx`.

We currently have `1/x^3dx` in the integrand, so we need to multiply by `-2` in the integral and `-1/2` to balance this on the exterior of the integral.

`=inte^(1//x^2)(1/x^3dx)=-1/2inte^(1//x^2)(-2/x^3dx)=-1/2inte^udu`

The integral of `e^u` is itself:

`=-1/2e^u=-e^(1//x^2)/2+C`