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

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

Integreer termgewijs:

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

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

$$- \int{\frac{1}{4 x^{4}} d x} + {\color{red}{\int{x^{4} d x}}}=- \int{\frac{1}{4 x^{4}} d x} + {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=- \int{\frac{1}{4 x^{4}} d x} + {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(x \right)} = \frac{1}{x^{4}}$$$:

$$\frac{x^{5}}{5} - {\color{red}{\int{\frac{1}{4 x^{4}} d x}}} = \frac{x^{5}}{5} - {\color{red}{\left(\frac{\int{\frac{1}{x^{4}} d x}}{4}\right)}}$$

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

$$\frac{x^{5}}{5} - \frac{{\color{red}{\int{\frac{1}{x^{4}} d x}}}}{4}=\frac{x^{5}}{5} - \frac{{\color{red}{\int{x^{-4} d x}}}}{4}=\frac{x^{5}}{5} - \frac{{\color{red}{\frac{x^{-4 + 1}}{-4 + 1}}}}{4}=\frac{x^{5}}{5} - \frac{{\color{red}{\left(- \frac{x^{-3}}{3}\right)}}}{4}=\frac{x^{5}}{5} - \frac{{\color{red}{\left(- \frac{1}{3 x^{3}}\right)}}}{4}$$

Dus,

$$\int{\left(x^{4} - \frac{1}{4 x^{4}}\right)d x} = \frac{x^{5}}{5} + \frac{1}{12 x^{3}}$$

Vereenvoudig:

$$\int{\left(x^{4} - \frac{1}{4 x^{4}}\right)d x} = \frac{12 x^{8} + 5}{60 x^{3}}$$

Voeg de integratieconstante toe:

$$\int{\left(x^{4} - \frac{1}{4 x^{4}}\right)d x} = \frac{12 x^{8} + 5}{60 x^{3}}+C$$

$$$\int \left(x^{4} - \frac{1}{4 x^{4}}\right)\, dx = \frac{12 x^{8} + 5}{60 x^{3}} + C$$$A