$$$\sqrt{x} \left(x - 1\right)$$$の積分

$$$\int \sqrt{x} \left(x - 1\right)\, dx$$$ を求めよ。

Expand the expression:

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

項別に積分せよ:

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

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

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

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

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

したがって、

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

簡単化せよ:

$$\int{\sqrt{x} \left(x - 1\right) d x} = \frac{2 x^{\frac{3}{2}} \left(3 x - 5\right)}{15}$$

積分定数を加える:

$$\int{\sqrt{x} \left(x - 1\right) d x} = \frac{2 x^{\frac{3}{2}} \left(3 x - 5\right)}{15}+C$$

$$$\int \sqrt{x} \left(x - 1\right)\, dx = \frac{2 x^{\frac{3}{2}} \left(3 x - 5\right)}{15} + C$$$A