Integralen av $$$\frac{x - 1}{x + 1}$$$

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

Låt $$$u=x + 1$$$ vara.

Då $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

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

Expand the expression:

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

Integrera termvis:

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

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=1$$$:

$$- \int{\frac{2}{u} d u} + {\color{red}{\int{1 d u}}} = - \int{\frac{2}{u} d u} + {\color{red}{u}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=2$$$ och $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

Integralen av $$$\frac{1}{u}$$$ är $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$u - 2 {\color{red}{\int{\frac{1}{u} d u}}} = u - 2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Kom ihåg att $$$u=x + 1$$$:

$$- 2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + {\color{red}{u}} = - 2 \ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)} + {\color{red}{\left(x + 1\right)}}$$

Alltså,

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

Lägg till integrationskonstanten (och ta bort konstanten från uttrycket):

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

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