Funktion $$$\frac{\sqrt{2 - x}}{2 x}$$$ integraali

Määritä $$$\int \frac{\sqrt{2 - x}}{2 x}\, dx$$$.

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

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

Olkoon $$$u=\sqrt{2 - x}$$$.

Tällöin $$$du=\left(\sqrt{2 - x}\right)^{\prime }dx = - \frac{1}{2 \sqrt{2 - x}} dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\frac{dx}{\sqrt{2 - x}} = - 2 du$$$.

Integraali voidaan kirjoittaa muotoon

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=-2$$$ ja $$$f{\left(u \right)} = \frac{u^{2}}{2 - u^{2}}$$$:

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

Koska osoittajan aste ei ole pienempi kuin nimittäjän aste, suorita polynomien jakokulma (vaiheet voidaan nähdä »):

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

Integroi termi kerrallaan:

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

Sovella vakiosääntöä $$$\int c\, du = c u$$$ käyttäen $$$c=1$$$:

$$- \int{\frac{2}{2 - u^{2}} d u} + {\color{red}{\int{1 d u}}} = - \int{\frac{2}{2 - u^{2}} d u} + {\color{red}{u}}$$

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

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

Suorita osamurtokehittely (vaiheet voidaan nähdä kohdassa »):

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

Integroi termi kerrallaan:

$$u - 2 {\color{red}{\int{\left(\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} - \frac{\sqrt{2}}{4 \left(u - \sqrt{2}\right)}\right)d u}}} = u - 2 {\color{red}{\left(- \int{\frac{\sqrt{2}}{4 \left(u - \sqrt{2}\right)} d u} + \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u}\right)}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{\sqrt{2}}{4}$$$ ja $$$f{\left(u \right)} = \frac{1}{u - \sqrt{2}}$$$:

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

Olkoon $$$v=u - \sqrt{2}$$$.

Tällöin $$$dv=\left(u - \sqrt{2}\right)^{\prime }du = 1 du$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$du = dv$$$.

Siis,

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

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

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

Muista, että $$$v=u - \sqrt{2}$$$:

$$u + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u} = u + \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(u - \sqrt{2}\right)}}}\right| \right)}}{2} - 2 \int{\frac{\sqrt{2}}{4 \left(u + \sqrt{2}\right)} d u}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{\sqrt{2}}{4}$$$ ja $$$f{\left(u \right)} = \frac{1}{u + \sqrt{2}}$$$:

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

Olkoon $$$v=u + \sqrt{2}$$$.

Tällöin $$$dv=\left(u + \sqrt{2}\right)^{\prime }du = 1 du$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$du = dv$$$.

Näin ollen,

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

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

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

Muista, että $$$v=u + \sqrt{2}$$$:

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

Muista, että $$$u=\sqrt{2 - x}$$$:

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

Näin ollen,

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

Lisää integrointivakio:

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

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