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

Bestäm $$$\int \frac{1 - x^{2}}{1 - x}\, dx$$$.

Förenkla integranden:

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

Integrera termvis:

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

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

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

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

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

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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