Integraali $$$\frac{1}{- k^{2} + r^{2}}$$$:stä muuttujan $$$k$$$ suhteen

Määritä $$$\int \frac{1}{- k^{2} + r^{2}}\, dk$$$.

Hajota osamurtoihin:

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

Integroi termi kerrallaan:

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

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

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

Olkoon $$$u=k + r$$$.

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

Siis,

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

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

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

Muista, että $$$u=k + r$$$:

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(k \right)}\, dk = c \int f{\left(k \right)}\, dk$$$ käyttäen $$$c=\frac{1}{2 r}$$$ ja $$$f{\left(k \right)} = \frac{1}{- k + r}$$$:

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

Olkoon $$$u=- k + r$$$.

Tällöin $$$du=\left(- k + r\right)^{\prime }dk = - dk$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dk = - du$$$.

Näin ollen,

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

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

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

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

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

Muista, että $$$u=- k + r$$$:

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

Näin ollen,

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

Sievennä:

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

Lisää integrointivakio:

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

$$$\int \frac{1}{- k^{2} + r^{2}}\, dk = \frac{- \ln\left(\left|{k - r}\right|\right) + \ln\left(\left|{k + r}\right|\right)}{2 r} + C$$$A