Integraal van $$$f \left(x + \frac{1}{x}\right)$$$ met betrekking tot $$$x$$$

Bepaal $$$\int f \left(x + \frac{1}{x}\right)\, dx$$$.

Expand the expression:

$${\color{red}{\int{f \left(x + \frac{1}{x}\right) d x}}} = {\color{red}{\int{\left(f x + \frac{f}{x}\right)d x}}}$$

Integreer termgewijs:

$${\color{red}{\int{\left(f x + \frac{f}{x}\right)d x}}} = {\color{red}{\left(\int{\frac{f}{x} d x} + \int{f x d x}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=f$$$ en $$$f{\left(x \right)} = x$$$:

$$\int{\frac{f}{x} d x} + {\color{red}{\int{f x d x}}} = \int{\frac{f}{x} d x} + {\color{red}{f \int{x d x}}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=1$$$:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=f$$$ en $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$\frac{f x^{2}}{2} + {\color{red}{\int{\frac{f}{x} d x}}} = \frac{f x^{2}}{2} + {\color{red}{f \int{\frac{1}{x} d x}}}$$

De integraal van $$$\frac{1}{x}$$$ is $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

Dus,

$$\int{f \left(x + \frac{1}{x}\right) d x} = \frac{f x^{2}}{2} + f \ln{\left(\left|{x}\right| \right)}$$

Vereenvoudig:

$$\int{f \left(x + \frac{1}{x}\right) d x} = \frac{f \left(x^{2} + 2 \ln{\left(\left|{x}\right| \right)}\right)}{2}$$

Voeg de integratieconstante toe:

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

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