$$$\frac{2}{7 - x^{2}}$$$의 적분

$$$\int \frac{2}{7 - x^{2}}\, dx$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=2$$$와 $$$f{\left(x \right)} = \frac{1}{7 - x^{2}}$$$에 적용하세요:

$${\color{red}{\int{\frac{2}{7 - x^{2}} d x}}} = {\color{red}{\left(2 \int{\frac{1}{7 - x^{2}} d x}\right)}}$$

부분분수분해를 수행합니다(단계는 »에서 볼 수 있습니다):

$$2 {\color{red}{\int{\frac{1}{7 - x^{2}} d x}}} = 2 {\color{red}{\int{\left(\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} - \frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)}\right)d x}}}$$

각 항별로 적분하십시오:

$$2 {\color{red}{\int{\left(\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} - \frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)}\right)d x}}} = 2 {\color{red}{\left(- \int{\frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)} d x} + \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}\right)}}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{\sqrt{7}}{14}$$$와 $$$f{\left(x \right)} = \frac{1}{x - \sqrt{7}}$$$에 적용하세요:

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - 2 {\color{red}{\int{\frac{\sqrt{7}}{14 \left(x - \sqrt{7}\right)} d x}}} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - 2 {\color{red}{\left(\frac{\sqrt{7} \int{\frac{1}{x - \sqrt{7}} d x}}{14}\right)}}$$

$$$u=x - \sqrt{7}$$$라 하자.

그러면 $$$du=\left(x - \sqrt{7}\right)^{\prime }dx = 1 dx$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dx = du$$$임을 얻습니다.

따라서,

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{x - \sqrt{7}} d x}}}}{7} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7}$$

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

$$2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7} = 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} - \frac{\sqrt{7} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{7}$$

다음 $$$u=x - \sqrt{7}$$$을 기억하라:

$$- \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{7} + 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x} = - \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{\left(x - \sqrt{7}\right)}}}\right| \right)}}{7} + 2 \int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{\sqrt{7}}{14}$$$와 $$$f{\left(x \right)} = \frac{1}{x + \sqrt{7}}$$$에 적용하세요:

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + 2 {\color{red}{\int{\frac{\sqrt{7}}{14 \left(x + \sqrt{7}\right)} d x}}} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + 2 {\color{red}{\left(\frac{\sqrt{7} \int{\frac{1}{x + \sqrt{7}} d x}}{14}\right)}}$$

$$$u=x + \sqrt{7}$$$라 하자.

그러면 $$$du=\left(x + \sqrt{7}\right)^{\prime }dx = 1 dx$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dx = du$$$임을 얻습니다.

적분은 다음과 같이 다시 쓸 수 있습니다.

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{x + \sqrt{7}} d x}}}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7}$$

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

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\int{\frac{1}{u} d u}}}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{7}$$

다음 $$$u=x + \sqrt{7}$$$을 기억하라:

$$- \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{7} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{{\color{red}{\left(x + \sqrt{7}\right)}}}\right| \right)}}{7}$$

따라서,

$$\int{\frac{2}{7 - x^{2}} d x} = - \frac{\sqrt{7} \ln{\left(\left|{x - \sqrt{7}}\right| \right)}}{7} + \frac{\sqrt{7} \ln{\left(\left|{x + \sqrt{7}}\right| \right)}}{7}$$

간단히 하시오:

$$\int{\frac{2}{7 - x^{2}} d x} = \frac{\sqrt{7} \left(- \ln{\left(\left|{x - \sqrt{7}}\right| \right)} + \ln{\left(\left|{x + \sqrt{7}}\right| \right)}\right)}{7}$$

적분 상수를 추가하세요:

$$\int{\frac{2}{7 - x^{2}} d x} = \frac{\sqrt{7} \left(- \ln{\left(\left|{x - \sqrt{7}}\right| \right)} + \ln{\left(\left|{x + \sqrt{7}}\right| \right)}\right)}{7}+C$$

$$$\int \frac{2}{7 - x^{2}}\, dx = \frac{\sqrt{7} \left(- \ln\left(\left|{x - \sqrt{7}}\right|\right) + \ln\left(\left|{x + \sqrt{7}}\right|\right)\right)}{7} + C$$$A