Integraal van $$$- x^{2} + \frac{1}{u}$$$ met betrekking tot $$$x$$$

Bepaal $$$\int \left(- x^{2} + \frac{1}{u}\right)\, dx$$$.

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=\frac{1}{u}$$$:

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \left(- x^{2} + \frac{1}{u}\right)\, dx = \left(- \frac{x^{3}}{3} + \frac{x}{u}\right) + C$$$A