$$$\frac{a - x}{\sqrt{x}}$$$ の $$$x$$$ に関する積分

$$$\int \frac{a - x}{\sqrt{x}}\, dx$$$ を求めよ。

Expand the expression:

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

項別に積分せよ:

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

$$$n=\frac{1}{2}$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=a$$$ と $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$ に対して適用する:

$$- \frac{2 x^{\frac{3}{2}}}{3} + {\color{red}{\int{\frac{a}{\sqrt{x}} d x}}} = - \frac{2 x^{\frac{3}{2}}}{3} + {\color{red}{a \int{\frac{1}{\sqrt{x}} d x}}}$$

$$$n=- \frac{1}{2}$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

したがって、

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

簡単化せよ:

$$\int{\frac{a - x}{\sqrt{x}} d x} = \frac{2 \sqrt{x} \left(3 a - x\right)}{3}$$

積分定数を加える:

$$\int{\frac{a - x}{\sqrt{x}} d x} = \frac{2 \sqrt{x} \left(3 a - x\right)}{3}+C$$

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