Integralen av $$$\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)}$$$

Bestäm $$$\int \frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)}\, dx$$$.

Utför partialbråksuppdelning (stegen kan ses »):

$${\color{red}{\int{\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)} d x}}} = {\color{red}{\int{\left(- \frac{1}{x - 2} + \frac{5}{3 \left(x - 3\right)} - \frac{2}{3 x}\right)d x}}}$$

Integrera termvis:

$${\color{red}{\int{\left(- \frac{1}{x - 2} + \frac{5}{3 \left(x - 3\right)} - \frac{2}{3 x}\right)d x}}} = {\color{red}{\left(- \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - \int{\frac{1}{x - 2} d x}\right)}}$$

Låt $$$u=x - 2$$$ vara.

Då $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Integralen blir

$$- \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{1}{x - 2} d x}}} = - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$- \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{1}{u} d u}}} = - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Kom ihåg att $$$u=x - 2$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x} = - \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)} - \int{\frac{2}{3 x} d x} + \int{\frac{5}{3 \left(x - 3\right)} d x}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{2}{3}$$$ och $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\int{\frac{2}{3 x} d x}}} = - \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - {\color{red}{\left(\frac{2 \int{\frac{1}{x} d x}}{3}\right)}}$$

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

$$- \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - \frac{2 {\color{red}{\int{\frac{1}{x} d x}}}}{3} = - \ln{\left(\left|{x - 2}\right| \right)} + \int{\frac{5}{3 \left(x - 3\right)} d x} - \frac{2 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{3}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{5}{3}$$$ och $$$f{\left(x \right)} = \frac{1}{x - 3}$$$:

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + {\color{red}{\int{\frac{5}{3 \left(x - 3\right)} d x}}} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + {\color{red}{\left(\frac{5 \int{\frac{1}{x - 3} d x}}{3}\right)}}$$

Låt $$$u=x - 3$$$ vara.

Då $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Integralen kan omskrivas som

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\int{\frac{1}{x - 3} d x}}}}{3} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{3}$$

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

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{3} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{3}$$

Kom ihåg att $$$u=x - 3$$$:

$$- \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{3} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)} + \frac{5 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)}}{3}$$

Alltså,

$$\int{\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)} d x} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} + \frac{5 \ln{\left(\left|{x - 3}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)}$$

Lägg till integrationskonstanten:

$$\int{\frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)} d x} = - \frac{2 \ln{\left(\left|{x}\right| \right)}}{3} + \frac{5 \ln{\left(\left|{x - 3}\right| \right)}}{3} - \ln{\left(\left|{x - 2}\right| \right)}+C$$

$$$\int \frac{3 x - 4}{x \left(x - 3\right) \left(x - 2\right)}\, dx = \left(- \frac{2 \ln\left(\left|{x}\right|\right)}{3} + \frac{5 \ln\left(\left|{x - 3}\right|\right)}{3} - \ln\left(\left|{x - 2}\right|\right)\right) + C$$$A