Funktion $$$\frac{5}{1 - x^{2}}$$$ integraali

Määritä $$$\int \frac{5}{1 - x^{2}}\, dx$$$.

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=5$$$ ja $$$f{\left(x \right)} = \frac{1}{1 - x^{2}}$$$:

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

Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):

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

Integroi termi kerrallaan:

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

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

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

Olkoon $$$u=x + 1$$$.

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

Integraali muuttuu muotoon

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

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

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

Muista, että $$$u=x + 1$$$:

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

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

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

Olkoon $$$u=x - 1$$$.

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

Integraali muuttuu muotoon

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

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

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

Muista, että $$$u=x - 1$$$:

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

Näin ollen,

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

Sievennä:

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

Lisää integrointivakio:

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

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