Funktion $$$\frac{1}{x^{4} - 1}$$$ integraali

Määritä $$$\int \frac{1}{x^{4} - 1}\, dx$$$.

Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):

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

Integroi termi kerrallaan:

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

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

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

Funktion $$$\frac{1}{x^{2} + 1}$$$ integraali on $$$\int{\frac{1}{x^{2} + 1} d x} = \operatorname{atan}{\left(x \right)}$$$:

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{4}$$$ ja $$$f{\left(x \right)} = \frac{1}{x + 1}$$$:

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

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

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

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

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

Muista, että $$$u=x + 1$$$:

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{4}$$$ ja $$$f{\left(x \right)} = \frac{1}{x - 1}$$$:

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

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

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

Integraali voidaan kirjoittaa muotoon

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

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

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

Muista, että $$$u=x - 1$$$:

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

Näin ollen,

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

Lisää integrointivakio:

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

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