Funktion $$$\frac{1}{1 - y^{2}}$$$ integraali

Määritä $$$\int \frac{1}{1 - y^{2}}\, dy$$$.

Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):

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

Integroi termi kerrallaan:

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(y \right)} = \frac{1}{y + 1}$$$:

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

Olkoon $$$u=y + 1$$$.

Tällöin $$$du=\left(y + 1\right)^{\prime }dy = 1 dy$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dy = du$$$.

Näin ollen,

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

Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Muista, että $$$u=y + 1$$$:

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(y \right)} = \frac{1}{y - 1}$$$:

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

Olkoon $$$u=y - 1$$$.

Tällöin $$$du=\left(y - 1\right)^{\prime }dy = 1 dy$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dy = du$$$.

Siis,

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

Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Muista, että $$$u=y - 1$$$:

$$\frac{\ln{\left(\left|{y + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{y + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(y - 1\right)}}}\right| \right)}}{2}$$

Näin ollen,

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

Sievennä:

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

Lisää integrointivakio:

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

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