Ολοκλήρωμα του $$$\frac{x}{\left(x - 1\right)^{2}}$$$

Βρείτε $$$\int \frac{x}{\left(x - 1\right)^{2}}\, dx$$$.

Ξαναγράψτε τον αριθμητή του ολοκληρωτέου ως $$$x=x - 1+1$$$ και διασπάστε το κλάσμα:

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

Ολοκληρώστε όρο προς όρο:

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

Έστω $$$u=x - 1$$$.

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

Επομένως,

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

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

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

Θυμηθείτε ότι $$$u=x - 1$$$:

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

Έστω $$$u=x - 1$$$.

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

Επομένως,

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

Εφαρμόστε τον κανόνα δύναμης $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ με $$$n=-2$$$:

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

Θυμηθείτε ότι $$$u=x - 1$$$:

$$\ln{\left(\left|{x - 1}\right| \right)} - {\color{red}{u}}^{-1} = \ln{\left(\left|{x - 1}\right| \right)} - {\color{red}{\left(x - 1\right)}}^{-1}$$

Επομένως,

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

Απλοποιήστε:

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

Προσθέστε τη σταθερά ολοκλήρωσης:

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

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