$$$x^{\frac{3}{2}} - \frac{3}{\sqrt{x}}$$$の積分

$$$\int \left(x^{\frac{3}{2}} - \frac{3}{\sqrt{x}}\right)\, dx$$$ を求めよ。

項別に積分せよ:

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

$$$n=\frac{3}{2}$$$ 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:

$$- \int{\frac{3}{\sqrt{x}} d x} + {\color{red}{\int{x^{\frac{3}{2}} d x}}}=- \int{\frac{3}{\sqrt{x}} d x} + {\color{red}{\frac{x^{1 + \frac{3}{2}}}{1 + \frac{3}{2}}}}=- \int{\frac{3}{\sqrt{x}} d x} + {\color{red}{\left(\frac{2 x^{\frac{5}{2}}}{5}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=3$$$ と $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$ に対して適用する:

$$\frac{2 x^{\frac{5}{2}}}{5} - {\color{red}{\int{\frac{3}{\sqrt{x}} d x}}} = \frac{2 x^{\frac{5}{2}}}{5} - {\color{red}{\left(3 \int{\frac{1}{\sqrt{x}} d x}\right)}}$$

$$$n=- \frac{1}{2}$$$ 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:

$$\frac{2 x^{\frac{5}{2}}}{5} - 3 {\color{red}{\int{\frac{1}{\sqrt{x}} d x}}}=\frac{2 x^{\frac{5}{2}}}{5} - 3 {\color{red}{\int{x^{- \frac{1}{2}} d x}}}=\frac{2 x^{\frac{5}{2}}}{5} - 3 {\color{red}{\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}=\frac{2 x^{\frac{5}{2}}}{5} - 3 {\color{red}{\left(2 x^{\frac{1}{2}}\right)}}=\frac{2 x^{\frac{5}{2}}}{5} - 3 {\color{red}{\left(2 \sqrt{x}\right)}}$$

したがって、

$$\int{\left(x^{\frac{3}{2}} - \frac{3}{\sqrt{x}}\right)d x} = \frac{2 x^{\frac{5}{2}}}{5} - 6 \sqrt{x}$$

簡単化せよ:

$$\int{\left(x^{\frac{3}{2}} - \frac{3}{\sqrt{x}}\right)d x} = \frac{2 \sqrt{x} \left(x^{2} - 15\right)}{5}$$

積分定数を加える:

$$\int{\left(x^{\frac{3}{2}} - \frac{3}{\sqrt{x}}\right)d x} = \frac{2 \sqrt{x} \left(x^{2} - 15\right)}{5}+C$$

$$$\int \left(x^{\frac{3}{2}} - \frac{3}{\sqrt{x}}\right)\, dx = \frac{2 \sqrt{x} \left(x^{2} - 15\right)}{5} + C$$$A