$$$k$$$ değişkenine göre $$$\frac{1}{- k^{2} + r^{2}}$$$ fonksiyonunun integrali

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

Kısmi kesirlere ayırın:

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

Her terimin integralini alın:

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

Sabit katsayı kuralı $$$\int c f{\left(k \right)}\, dk = c \int f{\left(k \right)}\, dk$$$'i $$$c=\frac{1}{2 r}$$$ ve $$$f{\left(k \right)} = \frac{1}{k + r}$$$ ile uygula:

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

$$$u=k + r$$$ olsun.

Böylece $$$du=\left(k + r\right)^{\prime }dk = 1 dk$$$ (adımlar » görülebilir) ve $$$dk = du$$$ elde ederiz.

İntegral şu şekilde yeniden yazılabilir:

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

$$$\frac{1}{u}$$$'nin integrali $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Hatırlayın ki $$$u=k + r$$$:

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

Sabit katsayı kuralı $$$\int c f{\left(k \right)}\, dk = c \int f{\left(k \right)}\, dk$$$'i $$$c=\frac{1}{2 r}$$$ ve $$$f{\left(k \right)} = \frac{1}{- k + r}$$$ ile uygula:

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

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

Böylece $$$du=\left(- k + r\right)^{\prime }dk = - dk$$$ (adımlar » görülebilir) ve $$$dk = - du$$$ elde ederiz.

O halde,

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

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=-1$$$ ve $$$f{\left(u \right)} = \frac{1}{u}$$$ ile uygula:

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

$$$\frac{1}{u}$$$'nin integrali $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Hatırlayın ki $$$u=- k + r$$$:

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

Dolayısıyla,

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

Sadeleştirin:

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

İntegrasyon sabitini ekleyin:

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

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