Integraal van $$$\frac{x - 1}{x^{3}}$$$

Bepaal $$$\int \frac{x - 1}{x^{3}}\, dx$$$.

Expand the expression:

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

Integreer termgewijs:

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

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

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

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-3$$$:

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

Dus,

$$\int{\frac{x - 1}{x^{3}} d x} = - \frac{1}{x} + \frac{1}{2 x^{2}}$$

Vereenvoudig:

$$\int{\frac{x - 1}{x^{3}} d x} = \frac{\frac{1}{2} - x}{x^{2}}$$

Voeg de integratieconstante toe:

$$\int{\frac{x - 1}{x^{3}} d x} = \frac{\frac{1}{2} - x}{x^{2}}+C$$

$$$\int \frac{x - 1}{x^{3}}\, dx = \frac{\frac{1}{2} - x}{x^{2}} + C$$$A