Integralen av $$$\frac{x + 6}{x - 6}$$$

Bestäm $$$\int \frac{x + 6}{x - 6}\, dx$$$.

Låt $$$u=x - 6$$$ vara.

Då $$$du=\left(x - 6\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Integralen blir

$${\color{red}{\int{\frac{x + 6}{x - 6} d x}}} = {\color{red}{\int{\frac{u + 12}{u} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{u + 12}{u} d u}}} = {\color{red}{\int{\left(1 + \frac{12}{u}\right)d u}}}$$

Integrera termvis:

$${\color{red}{\int{\left(1 + \frac{12}{u}\right)d u}}} = {\color{red}{\left(\int{1 d u} + \int{\frac{12}{u} d u}\right)}}$$

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

$$\int{\frac{12}{u} d u} + {\color{red}{\int{1 d u}}} = \int{\frac{12}{u} d u} + {\color{red}{u}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=12$$$ och $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$u + {\color{red}{\int{\frac{12}{u} d u}}} = u + {\color{red}{\left(12 \int{\frac{1}{u} d u}\right)}}$$

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

$$u + 12 {\color{red}{\int{\frac{1}{u} d u}}} = u + 12 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Kom ihåg att $$$u=x - 6$$$:

$$12 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + {\color{red}{u}} = 12 \ln{\left(\left|{{\color{red}{\left(x - 6\right)}}}\right| \right)} + {\color{red}{\left(x - 6\right)}}$$

Alltså,

$$\int{\frac{x + 6}{x - 6} d x} = x + 12 \ln{\left(\left|{x - 6}\right| \right)} - 6$$

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

$$\int{\frac{x + 6}{x - 6} d x} = x + 12 \ln{\left(\left|{x - 6}\right| \right)}+C$$

$$$\int \frac{x + 6}{x - 6}\, dx = \left(x + 12 \ln\left(\left|{x - 6}\right|\right)\right) + C$$$A