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

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

Change the base:

$${\color{red}{\int{\frac{2^{\frac{1}{x}}}{x^{2}} d x}}} = {\color{red}{\int{\frac{e^{\frac{\ln{\left(2 \right)}}{x}}}{x^{2}} d x}}}$$

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

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

Therefore,

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

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

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

Let $$$v=u \ln{\left(2 \right)}$$$.

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

The integral becomes

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

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

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

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

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

Recall that $$$v=u \ln{\left(2 \right)}$$$:

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

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

$$- \frac{e^{\ln{\left(2 \right)} {\color{red}{u}}}}{\ln{\left(2 \right)}} = - \frac{e^{\ln{\left(2 \right)} {\color{red}{\frac{1}{x}}}}}{\ln{\left(2 \right)}}$$

Therefore,

$$\int{\frac{2^{\frac{1}{x}}}{x^{2}} d x} = - \frac{e^{\frac{\ln{\left(2 \right)}}{x}}}{\ln{\left(2 \right)}}$$

Add the constant of integration:

$$\int{\frac{2^{\frac{1}{x}}}{x^{2}} d x} = - \frac{e^{\frac{\ln{\left(2 \right)}}{x}}}{\ln{\left(2 \right)}}+C$$

$$$\int \frac{2^{\frac{1}{x}}}{x^{2}}\, dx = - \frac{e^{\frac{\ln\left(2\right)}{x}}}{\ln\left(2\right)} + C$$$A