Integraal van $$$-4 + \frac{3}{x}$$$

Bepaal $$$\int \left(-4 + \frac{3}{x}\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(-4 + \frac{3}{x}\right)d x}}} = {\color{red}{\left(- \int{4 d x} + \int{\frac{3}{x} d x}\right)}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=4$$$:

$$\int{\frac{3}{x} d x} - {\color{red}{\int{4 d x}}} = \int{\frac{3}{x} d x} - {\color{red}{\left(4 x\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=3$$$ en $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- 4 x + {\color{red}{\int{\frac{3}{x} d x}}} = - 4 x + {\color{red}{\left(3 \int{\frac{1}{x} d x}\right)}}$$

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

$$- 4 x + 3 {\color{red}{\int{\frac{1}{x} d x}}} = - 4 x + 3 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Dus,

$$\int{\left(-4 + \frac{3}{x}\right)d x} = - 4 x + 3 \ln{\left(\left|{x}\right| \right)}$$

Voeg de integratieconstante toe:

$$\int{\left(-4 + \frac{3}{x}\right)d x} = - 4 x + 3 \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \left(-4 + \frac{3}{x}\right)\, dx = \left(- 4 x + 3 \ln\left(\left|{x}\right|\right)\right) + C$$$A