$$$\frac{1}{5 x^{3} - 5 x^{2}}$$$의 적분

$$$\int \frac{1}{5 x^{3} - 5 x^{2}}\, dx$$$을(를) 구하시오.

피적분함수를 단순화하세요.:

$${\color{red}{\int{\frac{1}{5 x^{3} - 5 x^{2}} d x}}} = {\color{red}{\int{\frac{1}{5 x^{2} \left(x - 1\right)} d x}}}$$

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{1}{5}$$$와 $$$f{\left(x \right)} = \frac{1}{x^{2} \left(x - 1\right)}$$$에 적용하세요:

$${\color{red}{\int{\frac{1}{5 x^{2} \left(x - 1\right)} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{x^{2} \left(x - 1\right)} d x}}{5}\right)}}$$

부분분수분해를 수행합니다(단계는 »에서 볼 수 있습니다):

$$\frac{{\color{red}{\int{\frac{1}{x^{2} \left(x - 1\right)} d x}}}}{5} = \frac{{\color{red}{\int{\left(\frac{1}{x - 1} - \frac{1}{x} - \frac{1}{x^{2}}\right)d x}}}}{5}$$

각 항별로 적분하십시오:

$$\frac{{\color{red}{\int{\left(\frac{1}{x - 1} - \frac{1}{x} - \frac{1}{x^{2}}\right)d x}}}}{5} = \frac{{\color{red}{\left(- \int{\frac{1}{x^{2}} d x} - \int{\frac{1}{x} d x} + \int{\frac{1}{x - 1} d x}\right)}}}{5}$$

$$$u=x - 1$$$라 하자.

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

따라서,

$$- \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{5} = - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{5}$$

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

$$- \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{5} = - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{5}$$

다음 $$$u=x - 1$$$을 기억하라:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5} = \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{\int{\frac{1}{x} d x}}{5}$$

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

$$\frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{5} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{\int{\frac{1}{x^{2}} d x}}{5} - \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{5}$$

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=-2$$$에 적용합니다:

$$- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\int{\frac{1}{x^{2}} d x}}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\int{x^{-2} d x}}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\left(- x^{-1}\right)}}}{5}=- \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} - \frac{{\color{red}{\left(- \frac{1}{x}\right)}}}{5}$$

따라서,

$$\int{\frac{1}{5 x^{3} - 5 x^{2}} d x} = - \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{\ln{\left(\left|{x - 1}\right| \right)}}{5} + \frac{1}{5 x}$$

간단히 하시오:

$$\int{\frac{1}{5 x^{3} - 5 x^{2}} d x} = \frac{x \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 1}\right| \right)}\right) + 1}{5 x}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{5 x^{3} - 5 x^{2}} d x} = \frac{x \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{x - 1}\right| \right)}\right) + 1}{5 x}+C$$

$$$\int \frac{1}{5 x^{3} - 5 x^{2}}\, dx = \frac{x \left(- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{x - 1}\right|\right)\right) + 1}{5 x} + C$$$A