Integralen av $$$1 + \frac{1}{x^{2}}$$$

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

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=1$$$:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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