Funktion $$$\frac{10}{100 - x^{2}}$$$ integraali
Aiheeseen liittyvä laskin: Määrättyjen ja epäoleellisten integraalien laskin
Syötteesi
Määritä $$$\int \frac{10}{100 - x^{2}}\, dx$$$.
Ratkaisu
Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=10$$$ ja $$$f{\left(x \right)} = \frac{1}{100 - x^{2}}$$$:
$${\color{red}{\int{\frac{10}{100 - x^{2}} d x}}} = {\color{red}{\left(10 \int{\frac{1}{100 - x^{2}} d x}\right)}}$$
Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):
$$10 {\color{red}{\int{\frac{1}{100 - x^{2}} d x}}} = 10 {\color{red}{\int{\left(\frac{1}{20 \left(x + 10\right)} - \frac{1}{20 \left(x - 10\right)}\right)d x}}}$$
Integroi termi kerrallaan:
$$10 {\color{red}{\int{\left(\frac{1}{20 \left(x + 10\right)} - \frac{1}{20 \left(x - 10\right)}\right)d x}}} = 10 {\color{red}{\left(- \int{\frac{1}{20 \left(x - 10\right)} d x} + \int{\frac{1}{20 \left(x + 10\right)} d x}\right)}}$$
Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{20}$$$ ja $$$f{\left(x \right)} = \frac{1}{x - 10}$$$:
$$10 \int{\frac{1}{20 \left(x + 10\right)} d x} - 10 {\color{red}{\int{\frac{1}{20 \left(x - 10\right)} d x}}} = 10 \int{\frac{1}{20 \left(x + 10\right)} d x} - 10 {\color{red}{\left(\frac{\int{\frac{1}{x - 10} d x}}{20}\right)}}$$
Olkoon $$$u=x - 10$$$.
Tällöin $$$du=\left(x - 10\right)^{\prime }dx = 1 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = du$$$.
Siis,
$$10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x - 10} d x}}}}{2} = 10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$
Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = 10 \int{\frac{1}{20 \left(x + 10\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
Muista, että $$$u=x - 10$$$:
$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + 10 \int{\frac{1}{20 \left(x + 10\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x - 10\right)}}}\right| \right)}}{2} + 10 \int{\frac{1}{20 \left(x + 10\right)} d x}$$
Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{20}$$$ ja $$$f{\left(x \right)} = \frac{1}{x + 10}$$$:
$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + 10 {\color{red}{\int{\frac{1}{20 \left(x + 10\right)} d x}}} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + 10 {\color{red}{\left(\frac{\int{\frac{1}{x + 10} d x}}{20}\right)}}$$
Olkoon $$$u=x + 10$$$.
Tällöin $$$du=\left(x + 10\right)^{\prime }dx = 1 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = du$$$.
Siis,
$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{x + 10} d x}}}}{2} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$
Funktion $$$\frac{1}{u}$$$ integraali on $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
Muista, että $$$u=x + 10$$$:
$$- \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(x + 10\right)}}}\right| \right)}}{2}$$
Näin ollen,
$$\int{\frac{10}{100 - x^{2}} d x} = - \frac{\ln{\left(\left|{x - 10}\right| \right)}}{2} + \frac{\ln{\left(\left|{x + 10}\right| \right)}}{2}$$
Sievennä:
$$\int{\frac{10}{100 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 10}\right| \right)} + \ln{\left(\left|{x + 10}\right| \right)}}{2}$$
Lisää integrointivakio:
$$\int{\frac{10}{100 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 10}\right| \right)} + \ln{\left(\left|{x + 10}\right| \right)}}{2}+C$$
Vastaus
$$$\int \frac{10}{100 - x^{2}}\, dx = \frac{- \ln\left(\left|{x - 10}\right|\right) + \ln\left(\left|{x + 10}\right|\right)}{2} + C$$$A