Ολοκλήρωμα του $$$\frac{1}{2 - x^{2}}$$$

Βρείτε $$$\int \frac{1}{2 - x^{2}}\, dx$$$.

Εκτελέστε αποσύνθεση σε μερικά κλάσματα (τα βήματα μπορούν να προβληθούν »):

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

Ολοκληρώστε όρο προς όρο:

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

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{\sqrt{2}}{4}$$$ και $$$f{\left(x \right)} = \frac{1}{x - \sqrt{2}}$$$:

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

Έστω $$$u=x - \sqrt{2}$$$.

Τότε $$$du=\left(x - \sqrt{2}\right)^{\prime }dx = 1 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = du$$$.

Επομένως,

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

Το ολοκλήρωμα του $$$\frac{1}{u}$$$ είναι $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Θυμηθείτε ότι $$$u=x - \sqrt{2}$$$:

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

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{\sqrt{2}}{4}$$$ και $$$f{\left(x \right)} = \frac{1}{x + \sqrt{2}}$$$:

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

Έστω $$$u=x + \sqrt{2}$$$.

Τότε $$$du=\left(x + \sqrt{2}\right)^{\prime }dx = 1 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = du$$$.

Επομένως,

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

Το ολοκλήρωμα του $$$\frac{1}{u}$$$ είναι $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Θυμηθείτε ότι $$$u=x + \sqrt{2}$$$:

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

Επομένως,

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

Απλοποιήστε:

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

Προσθέστε τη σταθερά ολοκλήρωσης:

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

$$$\int \frac{1}{2 - x^{2}}\, dx = \frac{\sqrt{2} \left(- \ln\left(\left|{x - \sqrt{2}}\right|\right) + \ln\left(\left|{x + \sqrt{2}}\right|\right)\right)}{4} + C$$$A