$$$x$$$에 대한 $$$\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}}$$$의 적분

$$$\int \frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}}\, dx$$$을(를) 구하시오.

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

$${\color{red}{\int{\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}} d x}}} = {\color{red}{\left(\frac{\sqrt{2} x_{1} \int{x^{2} d x}}{2 \sqrt{\pi} e^{\frac{1}{2}}}\right)}}$$

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

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

따라서,

$$\int{\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}} d x} = \frac{\sqrt{2} x^{3} x_{1}}{6 \sqrt{\pi} e^{\frac{1}{2}}}$$

적분 상수를 추가하세요:

$$\int{\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}} d x} = \frac{\sqrt{2} x^{3} x_{1}}{6 \sqrt{\pi} e^{\frac{1}{2}}}+C$$

$$$\int \frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}}\, dx = \frac{\sqrt{2} x^{3} x_{1}}{6 \sqrt{\pi} e^{\frac{1}{2}}} + C$$$A