Integralen av $$$- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}$$$

Bestäm $$$\int \left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=-24$$$ och $$$f{\left(x \right)} = \frac{x}{\left(x - 5\right) \left(x - 3\right)}$$$:

$${\color{red}{\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x}}} = {\color{red}{\left(- 24 \int{\frac{x}{\left(x - 5\right) \left(x - 3\right)} d x}\right)}}$$

Utför partialbråksuppdelning (stegen kan ses »):

$$- 24 {\color{red}{\int{\frac{x}{\left(x - 5\right) \left(x - 3\right)} d x}}} = - 24 {\color{red}{\int{\left(- \frac{3}{2 \left(x - 3\right)} + \frac{5}{2 \left(x - 5\right)}\right)d x}}}$$

Integrera termvis:

$$- 24 {\color{red}{\int{\left(- \frac{3}{2 \left(x - 3\right)} + \frac{5}{2 \left(x - 5\right)}\right)d x}}} = - 24 {\color{red}{\left(\int{\frac{5}{2 \left(x - 5\right)} d x} - \int{\frac{3}{2 \left(x - 3\right)} d x}\right)}}$$

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

$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 24 {\color{red}{\int{\frac{3}{2 \left(x - 3\right)} d x}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 24 {\color{red}{\left(\frac{3 \int{\frac{1}{x - 3} d x}}{2}\right)}}$$

Låt $$$u=x - 3$$$ vara.

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

Alltså,

$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{x - 3} d x}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$- 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\int{\frac{1}{u} d u}}} = - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} + 36 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Kom ihåg att $$$u=x - 3$$$:

$$36 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} - 24 \int{\frac{5}{2 \left(x - 5\right)} d x} = 36 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} - 24 \int{\frac{5}{2 \left(x - 5\right)} d x}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{5}{2}$$$ och $$$f{\left(x \right)} = \frac{1}{x - 5}$$$:

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 24 {\color{red}{\int{\frac{5}{2 \left(x - 5\right)} d x}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 24 {\color{red}{\left(\frac{5 \int{\frac{1}{x - 5} d x}}{2}\right)}}$$

Låt $$$u=x - 5$$$ vara.

Då $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{x - 5} d x}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\int{\frac{1}{u} d u}}} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Kom ihåg att $$$u=x - 5$$$:

$$36 \ln{\left(\left|{x - 3}\right| \right)} - 60 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 36 \ln{\left(\left|{x - 3}\right| \right)} - 60 \ln{\left(\left|{{\color{red}{\left(x - 5\right)}}}\right| \right)}$$

Alltså,

$$\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x} = - 60 \ln{\left(\left|{x - 5}\right| \right)} + 36 \ln{\left(\left|{x - 3}\right| \right)}$$

Lägg till integrationskonstanten:

$$\int{\left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)d x} = - 60 \ln{\left(\left|{x - 5}\right| \right)} + 36 \ln{\left(\left|{x - 3}\right| \right)}+C$$

$$$\int \left(- \frac{24 x}{\left(x - 5\right) \left(x - 3\right)}\right)\, dx = \left(- 60 \ln\left(\left|{x - 5}\right|\right) + 36 \ln\left(\left|{x - 3}\right|\right)\right) + C$$$A