Integraal van $$$\frac{e^{\frac{1}{x}}}{x^{2}}$$$

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

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

Dan $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\frac{dx}{x^{2}} = - du$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = e^{u}$$$:

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

De integraal van de exponentiële functie is $$$\int{e^{u} d u} = e^{u}$$$:

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

We herinneren eraan dat $$$u=\frac{1}{x}$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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