$$$\frac{1}{y^{2} - 3}$$$의 적분

$$$\int \frac{1}{y^{2} - 3}\, dy$$$을(를) 구하시오.

부분분수분해를 수행합니다(단계는 »에서 볼 수 있습니다):

$${\color{red}{\int{\frac{1}{y^{2} - 3} d y}}} = {\color{red}{\int{\left(- \frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} + \frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)}\right)d y}}}$$

각 항별로 적분하십시오:

$${\color{red}{\int{\left(- \frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} + \frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)}\right)d y}}} = {\color{red}{\left(\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \int{\frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} d y}\right)}}$$

상수배 법칙 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$을 $$$c=\frac{\sqrt{3}}{6}$$$와 $$$f{\left(y \right)} = \frac{1}{y + \sqrt{3}}$$$에 적용하세요:

$$\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - {\color{red}{\int{\frac{\sqrt{3}}{6 \left(y + \sqrt{3}\right)} d y}}} = \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{y + \sqrt{3}} d y}}{6}\right)}}$$

$$$u=y + \sqrt{3}$$$라 하자.

그러면 $$$du=\left(y + \sqrt{3}\right)^{\prime }dy = 1 dy$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dy = du$$$임을 얻습니다.

따라서,

$$\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{y + \sqrt{3}} d y}}}}{6} = \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

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

$$\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6} = \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} - \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

다음 $$$u=y + \sqrt{3}$$$을 기억하라:

$$- \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} + \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y} = - \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{\left(y + \sqrt{3}\right)}}}\right| \right)}}{6} + \int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y}$$

상수배 법칙 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$을 $$$c=\frac{\sqrt{3}}{6}$$$와 $$$f{\left(y \right)} = \frac{1}{y - \sqrt{3}}$$$에 적용하세요:

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + {\color{red}{\int{\frac{\sqrt{3}}{6 \left(y - \sqrt{3}\right)} d y}}} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{y - \sqrt{3}} d y}}{6}\right)}}$$

$$$u=y - \sqrt{3}$$$라 하자.

그러면 $$$du=\left(y - \sqrt{3}\right)^{\prime }dy = 1 dy$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dy = du$$$임을 얻습니다.

적분은 다음과 같이 됩니다.

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{y - \sqrt{3}} d y}}}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

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

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

다음 $$$u=y - \sqrt{3}$$$을 기억하라:

$$- \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{\left(y - \sqrt{3}\right)}}}\right| \right)}}{6}$$

따라서,

$$\int{\frac{1}{y^{2} - 3} d y} = \frac{\sqrt{3} \ln{\left(\left|{y - \sqrt{3}}\right| \right)}}{6} - \frac{\sqrt{3} \ln{\left(\left|{y + \sqrt{3}}\right| \right)}}{6}$$

간단히 하시오:

$$\int{\frac{1}{y^{2} - 3} d y} = \frac{\sqrt{3} \left(\ln{\left(\left|{y - \sqrt{3}}\right| \right)} - \ln{\left(\left|{y + \sqrt{3}}\right| \right)}\right)}{6}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{y^{2} - 3} d y} = \frac{\sqrt{3} \left(\ln{\left(\left|{y - \sqrt{3}}\right| \right)} - \ln{\left(\left|{y + \sqrt{3}}\right| \right)}\right)}{6}+C$$

$$$\int \frac{1}{y^{2} - 3}\, dy = \frac{\sqrt{3} \left(\ln\left(\left|{y - \sqrt{3}}\right|\right) - \ln\left(\left|{y + \sqrt{3}}\right|\right)\right)}{6} + C$$$A