Integraal van $$$- \frac{x}{e^{2}} + x e^{2}$$$

Bepaal $$$\int \left(- \frac{x}{e^{2}} + x e^{2}\right)\, dx$$$.

Integreer termgewijs:

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=e^{2}$$$ en $$$f{\left(x \right)} = x$$$:

$$- \int{\frac{x}{e^{2}} d x} + {\color{red}{\int{x e^{2} d x}}} = - \int{\frac{x}{e^{2}} d x} + {\color{red}{e^{2} \int{x d x}}}$$

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

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=e^{-2}$$$ en $$$f{\left(x \right)} = x$$$:

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

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

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

Dus,

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

Vereenvoudig:

$$\int{\left(- \frac{x}{e^{2}} + x e^{2}\right)d x} = x^{2} \sinh{\left(2 \right)}$$

Voeg de integratieconstante toe:

$$\int{\left(- \frac{x}{e^{2}} + x e^{2}\right)d x} = x^{2} \sinh{\left(2 \right)}+C$$

$$$\int \left(- \frac{x}{e^{2}} + x e^{2}\right)\, dx = x^{2} \sinh{\left(2 \right)} + C$$$A