$$$\frac{1}{t^{3} - t}$$$의 적분

$$$\int \frac{1}{t^{3} - t}\, dt$$$을(를) 구하시오.

부분분수분해를 수행합니다(단계는 »에서 볼 수 있습니다):

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

각 항별로 적분하십시오:

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

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

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

$$$u=t + 1$$$라 하자.

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

따라서,

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

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

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

다음 $$$u=t + 1$$$을 기억하라:

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

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

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

$$$u=t - 1$$$라 하자.

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

따라서,

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

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

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

다음 $$$u=t - 1$$$을 기억하라:

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

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

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

따라서,

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

적분 상수를 추가하세요:

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

$$$\int \frac{1}{t^{3} - t}\, dt = \left(- \ln\left(\left|{t}\right|\right) + \frac{\ln\left(\left|{t - 1}\right|\right)}{2} + \frac{\ln\left(\left|{t + 1}\right|\right)}{2}\right) + C$$$A