Integraal van $$$i f n t x + g x$$$ met betrekking tot $$$x$$$

Bepaal $$$\int \left(i f n t x + g x\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(i f n t x + g x\right)d x}}} = {\color{red}{\left(\int{g x d x} + \int{i f n t x d x}\right)}}$$

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

$$\int{i f n t x d x} + {\color{red}{\int{g x d x}}} = \int{i f n t x d x} + {\color{red}{g \int{x d x}}}$$

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

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

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

$$\frac{g x^{2}}{2} + {\color{red}{\int{i f n t x d x}}} = \frac{g x^{2}}{2} + {\color{red}{i f n t \int{x d x}}}$$

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

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

Dus,

$$\int{\left(i f n t x + g x\right)d x} = \frac{i f n t x^{2}}{2} + \frac{g x^{2}}{2}$$

Vereenvoudig:

$$\int{\left(i f n t x + g x\right)d x} = \frac{x^{2} \left(i f n t + g\right)}{2}$$

Voeg de integratieconstante toe:

$$\int{\left(i f n t x + g x\right)d x} = \frac{x^{2} \left(i f n t + g\right)}{2}+C$$

$$$\int \left(i f n t x + g x\right)\, dx = \frac{x^{2} \left(i f n t + g\right)}{2} + C$$$A