Integralen av $$$\frac{7}{2 x^{2} - x - 3}$$$

Bestäm $$$\int \frac{7}{2 x^{2} - x - 3}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=7$$$ och $$$f{\left(x \right)} = \frac{1}{2 x^{2} - x - 3}$$$:

$${\color{red}{\int{\frac{7}{2 x^{2} - x - 3} d x}}} = {\color{red}{\left(7 \int{\frac{1}{2 x^{2} - x - 3} d x}\right)}}$$

Utför partialbråksuppdelning (stegen kan ses »):

$$7 {\color{red}{\int{\frac{1}{2 x^{2} - x - 3} d x}}} = 7 {\color{red}{\int{\left(\frac{2}{5 \left(2 x - 3\right)} - \frac{1}{5 \left(x + 1\right)}\right)d x}}}$$

Integrera termvis:

$$7 {\color{red}{\int{\left(\frac{2}{5 \left(2 x - 3\right)} - \frac{1}{5 \left(x + 1\right)}\right)d x}}} = 7 {\color{red}{\left(- \int{\frac{1}{5 \left(x + 1\right)} d x} + \int{\frac{2}{5 \left(2 x - 3\right)} d x}\right)}}$$

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

$$7 \int{\frac{2}{5 \left(2 x - 3\right)} d x} - 7 {\color{red}{\int{\frac{1}{5 \left(x + 1\right)} d x}}} = 7 \int{\frac{2}{5 \left(2 x - 3\right)} d x} - 7 {\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{5}\right)}}$$

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,

$$7 \int{\frac{2}{5 \left(2 x - 3\right)} d x} - \frac{7 {\color{red}{\int{\frac{1}{x + 1} d x}}}}{5} = 7 \int{\frac{2}{5 \left(2 x - 3\right)} d x} - \frac{7 {\color{red}{\int{\frac{1}{u} d u}}}}{5}$$

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

$$7 \int{\frac{2}{5 \left(2 x - 3\right)} d x} - \frac{7 {\color{red}{\int{\frac{1}{u} d u}}}}{5} = 7 \int{\frac{2}{5 \left(2 x - 3\right)} d x} - \frac{7 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{5}$$

Kom ihåg att $$$u=x + 1$$$:

$$- \frac{7 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{5} + 7 \int{\frac{2}{5 \left(2 x - 3\right)} d x} = - \frac{7 \ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{5} + 7 \int{\frac{2}{5 \left(2 x - 3\right)} d x}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{2}{5}$$$ och $$$f{\left(x \right)} = \frac{1}{2 x - 3}$$$:

$$- \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + 7 {\color{red}{\int{\frac{2}{5 \left(2 x - 3\right)} d x}}} = - \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + 7 {\color{red}{\left(\frac{2 \int{\frac{1}{2 x - 3} d x}}{5}\right)}}$$

Låt $$$u=2 x - 3$$$ vara.

Då $$$du=\left(2 x - 3\right)^{\prime }dx = 2 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{2}$$$.

Alltså,

$$- \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{14 {\color{red}{\int{\frac{1}{2 x - 3} d x}}}}{5} = - \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{14 {\color{red}{\int{\frac{1}{2 u} d u}}}}{5}$$

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

$$- \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{14 {\color{red}{\int{\frac{1}{2 u} d u}}}}{5} = - \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{14 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}}{5}$$

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

$$- \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{7 {\color{red}{\int{\frac{1}{u} d u}}}}{5} = - \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{7 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{5}$$

Kom ihåg att $$$u=2 x - 3$$$:

$$- \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{7 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{5} = - \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{7 \ln{\left(\left|{{\color{red}{\left(2 x - 3\right)}}}\right| \right)}}{5}$$

Alltså,

$$\int{\frac{7}{2 x^{2} - x - 3} d x} = - \frac{7 \ln{\left(\left|{x + 1}\right| \right)}}{5} + \frac{7 \ln{\left(\left|{2 x - 3}\right| \right)}}{5}$$

Förenkla:

$$\int{\frac{7}{2 x^{2} - x - 3} d x} = \frac{7 \left(- \ln{\left(\left|{x + 1}\right| \right)} + \ln{\left(\left|{2 x - 3}\right| \right)}\right)}{5}$$

Lägg till integrationskonstanten:

$$\int{\frac{7}{2 x^{2} - x - 3} d x} = \frac{7 \left(- \ln{\left(\left|{x + 1}\right| \right)} + \ln{\left(\left|{2 x - 3}\right| \right)}\right)}{5}+C$$

$$$\int \frac{7}{2 x^{2} - x - 3}\, dx = \frac{7 \left(- \ln\left(\left|{x + 1}\right|\right) + \ln\left(\left|{2 x - 3}\right|\right)\right)}{5} + C$$$A