Integraal van $$$\left(x + 1\right) \left(x + 2\right)$$$

Bepaal $$$\int \left(x + 1\right) \left(x + 2\right)\, dx$$$.

Expand the expression:

$${\color{red}{\int{\left(x + 1\right) \left(x + 2\right) d x}}} = {\color{red}{\int{\left(x^{2} + 3 x + 2\right)d x}}}$$

Integreer termgewijs:

$${\color{red}{\int{\left(x^{2} + 3 x + 2\right)d x}}} = {\color{red}{\left(\int{2 d x} + \int{3 x d x} + \int{x^{2} d x}\right)}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=2$$$:

$$\int{3 x d x} + \int{x^{2} d x} + {\color{red}{\int{2 d x}}} = \int{3 x d x} + \int{x^{2} d x} + {\color{red}{\left(2 x\right)}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=2$$$:

$$2 x + \int{3 x d x} + {\color{red}{\int{x^{2} d x}}}=2 x + \int{3 x d x} + {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=2 x + \int{3 x d x} + {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

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

$$\frac{x^{3}}{3} + 2 x + {\color{red}{\int{3 x d x}}} = \frac{x^{3}}{3} + 2 x + {\color{red}{\left(3 \int{x d x}\right)}}$$

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

$$\frac{x^{3}}{3} + 2 x + 3 {\color{red}{\int{x d x}}}=\frac{x^{3}}{3} + 2 x + 3 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\frac{x^{3}}{3} + 2 x + 3 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Dus,

$$\int{\left(x + 1\right) \left(x + 2\right) d x} = \frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x$$

Vereenvoudig:

$$\int{\left(x + 1\right) \left(x + 2\right) d x} = \frac{x \left(2 x^{2} + 9 x + 12\right)}{6}$$

Voeg de integratieconstante toe:

$$\int{\left(x + 1\right) \left(x + 2\right) d x} = \frac{x \left(2 x^{2} + 9 x + 12\right)}{6}+C$$

$$$\int \left(x + 1\right) \left(x + 2\right)\, dx = \frac{x \left(2 x^{2} + 9 x + 12\right)}{6} + C$$$A