$$$r$$$에 대한 $$$\frac{1}{- k^{2} + r^{2}}$$$의 적분

$$$\int \frac{1}{- k^{2} + r^{2}}\, dr$$$을(를) 구하시오.

부분분수 분해 수행:

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

각 항별로 적분하십시오:

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

상수배 법칙 $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$을 $$$c=\frac{1}{2 \left|{k}\right|}$$$와 $$$f{\left(r \right)} = \frac{1}{- k + r}$$$에 적용하세요:

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

$$$u=- k + r$$$라 하자.

그러면 $$$du=\left(- k + r\right)^{\prime }dr = 1 dr$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dr = du$$$임을 얻습니다.

적분은 다음과 같이 다시 쓸 수 있습니다.

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

$$$\frac{1}{u}$$$의 적분은 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

다음 $$$u=- k + r$$$을 기억하라:

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

상수배 법칙 $$$\int c f{\left(r \right)}\, dr = c \int f{\left(r \right)}\, dr$$$을 $$$c=\frac{1}{2 \left|{k}\right|}$$$와 $$$f{\left(r \right)} = \frac{1}{k + r}$$$에 적용하세요:

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

$$$u=k + r$$$라 하자.

그러면 $$$du=\left(k + r\right)^{\prime }dr = 1 dr$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dr = du$$$임을 얻습니다.

따라서,

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

$$$\frac{1}{u}$$$의 적분은 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

다음 $$$u=k + r$$$을 기억하라:

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

따라서,

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

간단히 하시오:

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

적분 상수를 추가하세요:

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

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