# How do you evaluate the integral int e^(1/x)/x^2?

##### 1 Answer
Jan 28, 2017

$\int {e}^{\frac{1}{x}} / {x}^{2} \mathrm{dx} = - {e}^{\frac{1}{x}} + C$

#### Explanation:

Substitute: $t = \frac{1}{x}$, $\mathrm{dt} = - \frac{1}{x} ^ 2 \mathrm{dx}$

so we have:

$\int {e}^{\frac{1}{x}} / {x}^{2} \mathrm{dx} = - \int {e}^{t} \mathrm{dt} = - {e}^{t} + C$

and undoing the substitution:

$\int {e}^{\frac{1}{x}} / {x}^{2} \mathrm{dx} = - {e}^{\frac{1}{x}} + C$
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