$$$\sqrt{3} \sqrt{x}$$$의 적분

$$$\int \sqrt{3} \sqrt{x}\, dx$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\sqrt{3}$$$와 $$$f{\left(x \right)} = \sqrt{x}$$$에 적용하세요:

$${\color{red}{\int{\sqrt{3} \sqrt{x} d x}}} = {\color{red}{\sqrt{3} \int{\sqrt{x} d x}}}$$

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

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

따라서,

$$\int{\sqrt{3} \sqrt{x} d x} = \frac{2 \sqrt{3} x^{\frac{3}{2}}}{3}$$

적분 상수를 추가하세요:

$$\int{\sqrt{3} \sqrt{x} d x} = \frac{2 \sqrt{3} x^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{3} \sqrt{x}\, dx = \frac{2 \sqrt{3} x^{\frac{3}{2}}}{3} + C$$$A