Funktion $$$\frac{1}{3 \left(1 - x^{2}\right)}$$$ integraali

Määritä $$$\int \frac{1}{3 \left(1 - x^{2}\right)}\, dx$$$.

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{3}$$$ ja $$$f{\left(x \right)} = \frac{1}{1 - x^{2}}$$$:

$${\color{red}{\int{\frac{1}{3 \left(1 - x^{2}\right)} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{1 - x^{2}} d x}}{3}\right)}}$$

Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):

$$\frac{{\color{red}{\int{\frac{1}{1 - x^{2}} d x}}}}{3} = \frac{{\color{red}{\int{\left(\frac{1}{2 \left(x + 1\right)} - \frac{1}{2 \left(x - 1\right)}\right)d x}}}}{3}$$

Integroi termi kerrallaan:

$$\frac{{\color{red}{\int{\left(\frac{1}{2 \left(x + 1\right)} - \frac{1}{2 \left(x - 1\right)}\right)d x}}}}{3} = \frac{{\color{red}{\left(- \int{\frac{1}{2 \left(x - 1\right)} d x} + \int{\frac{1}{2 \left(x + 1\right)} d x}\right)}}}{3}$$

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

$$- \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{2 \left(x + 1\right)} d x}}}}{3} = - \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3} + \frac{{\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{2}\right)}}}{3}$$

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

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

Siis,

$$- \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{x + 1} d x}}}}{6} = - \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

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

$$- \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6} = - \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Muista, että $$$u=x + 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} - \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3} = \frac{\ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{6} - \frac{\int{\frac{1}{2 \left(x - 1\right)} d x}}{3}$$

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

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{{\color{red}{\int{\frac{1}{2 \left(x - 1\right)} d x}}}}{3} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{{\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{2}\right)}}}{3}$$

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

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

Näin ollen,

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{6} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

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

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{6} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Muista, että $$$u=x - 1$$$:

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{6} - \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{6}$$

Näin ollen,

$$\int{\frac{1}{3 \left(1 - x^{2}\right)} d x} = - \frac{\ln{\left(\left|{x - 1}\right| \right)}}{6} + \frac{\ln{\left(\left|{x + 1}\right| \right)}}{6}$$

Sievennä:

$$\int{\frac{1}{3 \left(1 - x^{2}\right)} d x} = \frac{- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}}{6}$$

Lisää integrointivakio:

$$\int{\frac{1}{3 \left(1 - x^{2}\right)} d x} = \frac{- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}}{6}+C$$

$$$\int \frac{1}{3 \left(1 - x^{2}\right)}\, dx = \frac{- \ln\left(\left|{x - 1}\right|\right) + \ln\left(\left|{x + 1}\right|\right)}{6} + C$$$A