Integraal van $$$1 + \frac{1}{x^{4}}$$$

Bepaal $$$\int \left(1 + \frac{1}{x^{4}}\right)\, dx$$$.

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

$$\int{\frac{1}{x^{4}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{4}} d x} + {\color{red}{x}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-4$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \left(1 + \frac{1}{x^{4}}\right)\, dx = \left(x - \frac{1}{3 x^{3}}\right) + C$$$A