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

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{\sqrt{2}}{4}$$$ och $$$f{\left(x \right)} = \frac{1}{x \left(x - 3\right)}$$$:

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

Utför partialbråksuppdelning (stegen kan ses »):

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

Integrera termvis:

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

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

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

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

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

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

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

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 blir

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

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

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

Kom ihåg att $$$u=x - 3$$$:

$$\frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{3}\right)}{4} = \frac{\sqrt{2} \left(- \frac{\ln{\left(\left|{x}\right| \right)}}{3} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)}}{3}\right)}{4}$$

Alltså,

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

Förenkla:

$$\int{\frac{\sqrt{2}}{4 x \left(x - 3\right)} d x} = \frac{\sqrt{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 3}\right| \right)}\right)}{12}$$

Lägg till integrationskonstanten:

$$\int{\frac{\sqrt{2}}{4 x \left(x - 3\right)} d x} = \frac{\sqrt{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 3}\right| \right)}\right)}{12}+C$$

$$$\int \frac{\sqrt{2}}{4 x \left(x - 3\right)}\, dx = \frac{\sqrt{2} \left(- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{x - 3}\right|\right)\right)}{12} + C$$$A