$$$\left(x^{2} + 1\right)^{2}$$$の積分

$$$\int \left(x^{2} + 1\right)^{2}\, dx$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{\left(x^{2} + 1\right)^{2} d x}}} = {\color{red}{\int{\left(x^{4} + 2 x^{2} + 1\right)d x}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(x^{4} + 2 x^{2} + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{2 x^{2} d x} + \int{x^{4} d x}\right)}}$$

$$$c=1$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$\int{2 x^{2} d x} + \int{x^{4} d x} + {\color{red}{\int{1 d x}}} = \int{2 x^{2} d x} + \int{x^{4} d x} + {\color{red}{x}}$$

$$$n=4$$$ 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:

$$x + \int{2 x^{2} d x} + {\color{red}{\int{x^{4} d x}}}=x + \int{2 x^{2} d x} + {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=x + \int{2 x^{2} d x} + {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

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:

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

$$$n=2$$$ 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:

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

したがって、

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

積分定数を加える:

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

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