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

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \frac{e^{- \frac{2}{x}}}{x^{2}}$$$:

$${\color{red}{\int{\frac{2 e^{- \frac{2}{x}}}{x^{2}} d x}}} = {\color{red}{\left(2 \int{\frac{e^{- \frac{2}{x}}}{x^{2}} d x}\right)}}$$

Let $$$u=- \frac{2}{x}$$$.

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

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$2 {\color{red}{\int{\frac{e^{u}}{2} d u}}} = 2 {\color{red}{\left(\frac{\int{e^{u} d u}}{2}\right)}}$$

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 $$$u=- \frac{2}{x}$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{\left(- \frac{2}{x}\right)}}}$$

Therefore,

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

Add the constant of integration:

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

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