$$$x \left(20 x - 10\right) + \sqrt{3}$$$の積分

$$$\int \left(x \left(20 x - 10\right) + \sqrt{3}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x}}} = {\color{red}{\left(\int{\sqrt{3} d x} + \int{x \left(20 x - 10\right) d x}\right)}}$$

$$$c=\sqrt{3}$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$\int{x \left(20 x - 10\right) d x} + {\color{red}{\int{\sqrt{3} d x}}} = \int{x \left(20 x - 10\right) d x} + {\color{red}{\sqrt{3} x}}$$

被積分関数を簡単化する:

$$\sqrt{3} x + {\color{red}{\int{x \left(20 x - 10\right) d x}}} = \sqrt{3} x + {\color{red}{\int{10 x \left(2 x - 1\right) d x}}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=10$$$ と $$$f{\left(x \right)} = x \left(2 x - 1\right)$$$ に対して適用する:

$$\sqrt{3} x + {\color{red}{\int{10 x \left(2 x - 1\right) d x}}} = \sqrt{3} x + {\color{red}{\left(10 \int{x \left(2 x - 1\right) d x}\right)}}$$

Expand the expression:

$$\sqrt{3} x + 10 {\color{red}{\int{x \left(2 x - 1\right) d x}}} = \sqrt{3} x + 10 {\color{red}{\int{\left(2 x^{2} - x\right)d x}}}$$

項別に積分せよ:

$$\sqrt{3} x + 10 {\color{red}{\int{\left(2 x^{2} - x\right)d x}}} = \sqrt{3} x + 10 {\color{red}{\left(- \int{x d x} + \int{2 x^{2} d x}\right)}}$$

$$$n=1$$$ 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:

$$\sqrt{3} x + 10 \int{2 x^{2} d x} - 10 {\color{red}{\int{x d x}}}=\sqrt{3} x + 10 \int{2 x^{2} d x} - 10 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\sqrt{3} x + 10 \int{2 x^{2} d x} - 10 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=2$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用する:

$$- 5 x^{2} + \sqrt{3} x + 10 {\color{red}{\int{2 x^{2} d x}}} = - 5 x^{2} + \sqrt{3} x + 10 {\color{red}{\left(2 \int{x^{2} d x}\right)}}$$

$$$n=2$$$ 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:

$$- 5 x^{2} + \sqrt{3} x + 20 {\color{red}{\int{x^{2} d x}}}=- 5 x^{2} + \sqrt{3} x + 20 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- 5 x^{2} + \sqrt{3} x + 20 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

したがって、

$$\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x} = \frac{20 x^{3}}{3} - 5 x^{2} + \sqrt{3} x$$

簡単化せよ:

$$\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x} = \frac{x \left(20 x^{2} - 15 x + 3 \sqrt{3}\right)}{3}$$

積分定数を加える:

$$\int{\left(x \left(20 x - 10\right) + \sqrt{3}\right)d x} = \frac{x \left(20 x^{2} - 15 x + 3 \sqrt{3}\right)}{3}+C$$

$$$\int \left(x \left(20 x - 10\right) + \sqrt{3}\right)\, dx = \frac{x \left(20 x^{2} - 15 x + 3 \sqrt{3}\right)}{3} + C$$$A