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

Jan 7, 2017

$- {e}^{1 / {x}^{2}} / 2 + C$

#### Explanation:

$\int {e}^{1 / {x}^{2}} / {x}^{3} \mathrm{dx}$

Let $u = \frac{1}{x} ^ 2$. Differentiating this, we see that $\mathrm{du} = - \frac{2}{x} ^ 3 \mathrm{dx}$.

We currently have $\frac{1}{x} ^ 3 \mathrm{dx}$ in the integrand, so we need to multiply by $- 2$ in the integral and $- \frac{1}{2}$ to balance this on the exterior of the integral.

$= \int {e}^{1 / {x}^{2}} \left(\frac{1}{x} ^ 3 \mathrm{dx}\right) = - \frac{1}{2} \int {e}^{1 / {x}^{2}} \left(- \frac{2}{x} ^ 3 \mathrm{dx}\right) = - \frac{1}{2} \int {e}^{u} \mathrm{du}$

The integral of ${e}^{u}$ is itself:

$= - \frac{1}{2} {e}^{u} = - {e}^{1 / {x}^{2}} / 2 + C$