$$$- \frac{6}{\sqrt{y^{3}}} + \frac{3}{\sqrt{y}}$$$의 적분

$$$\int \left(- \frac{6}{\sqrt{y^{3}}} + \frac{3}{\sqrt{y}}\right)\, dy$$$을(를) 구하시오.

입력이 다음과 같이 다시 쓰입니다: $$$\int{\left(- \frac{6}{\sqrt{y^{3}}} + \frac{3}{\sqrt{y}}\right)d y}=\int{\left(\frac{3}{\sqrt{y}} - \frac{6}{y^{\frac{3}{2}}}\right)d y}$$$.

각 항별로 적분하십시오:

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

상수배 법칙 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$을 $$$c=6$$$와 $$$f{\left(y \right)} = \frac{1}{y^{\frac{3}{2}}}$$$에 적용하세요:

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

멱법칙($$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=- \frac{3}{2}$$$에 적용합니다:

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

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

$${\color{red}{\int{\frac{3}{\sqrt{y}} d y}}} + \frac{12}{\sqrt{y}} = {\color{red}{\left(3 \int{\frac{1}{\sqrt{y}} d y}\right)}} + \frac{12}{\sqrt{y}}$$

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

$$3 {\color{red}{\int{\frac{1}{\sqrt{y}} d y}}} + \frac{12}{\sqrt{y}}=3 {\color{red}{\int{y^{- \frac{1}{2}} d y}}} + \frac{12}{\sqrt{y}}=3 {\color{red}{\frac{y^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}} + \frac{12}{\sqrt{y}}=3 {\color{red}{\left(2 y^{\frac{1}{2}}\right)}} + \frac{12}{\sqrt{y}}=3 {\color{red}{\left(2 \sqrt{y}\right)}} + \frac{12}{\sqrt{y}}$$

따라서,

$$\int{\left(\frac{3}{\sqrt{y}} - \frac{6}{y^{\frac{3}{2}}}\right)d y} = 6 \sqrt{y} + \frac{12}{\sqrt{y}}$$

간단히 하시오:

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

적분 상수를 추가하세요:

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

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