Integral of $$$\frac{e^{- \frac{1}{x}}}{x^{2}}$$$

Find $$$\int \frac{e^{- \frac{1}{x}}}{x^{2}}\, dx$$$.

Let $$$u=- \frac{1}{x}$$$.

Then $$$du=\left(- \frac{1}{x}\right)^{\prime }dx = \frac{dx}{x^{2}}$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{2}} = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{e^{- \frac{1}{x}}}{x^{2}} d x}}} = {\color{red}{\int{e^{u} d u}}}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$${\color{red}{\int{e^{u} d u}}} = {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{1}{x}$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{\left(- \frac{1}{x}\right)}}}$$

Therefore,

$$\int{\frac{e^{- \frac{1}{x}}}{x^{2}} d x} = e^{- \frac{1}{x}}$$

Add the constant of integration:

$$\int{\frac{e^{- \frac{1}{x}}}{x^{2}} d x} = e^{- \frac{1}{x}}+C$$

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