$$$- \sqrt{x} + \frac{1}{x}$$$의 적분

$$$\int \left(- \sqrt{x} + \frac{1}{x}\right)\, dx$$$을(를) 구하시오.

각 항별로 적분하십시오:

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

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

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

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

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

따라서,

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

적분 상수를 추가하세요:

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

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