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

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

Kısmi kesirlere ayırın:

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

Her terimin integralini alın:

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

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

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

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

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

İntegral şu şekilde yeniden yazılabilir:

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

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

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

Hatırlayın ki $$$u=- k + r$$$:

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

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

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

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

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

İntegral şu hale gelir

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

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

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

Hatırlayın ki $$$u=k + r$$$:

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

Dolayısıyla,

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

Sadeleştirin:

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

İntegrasyon sabitini ekleyin:

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

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