$$$z \left(1 + \frac{z}{t}\right)$$$ の $$$z$$$ に関する積分

$$$\int z \left(1 + \frac{z}{t}\right)\, dz$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{z \left(1 + \frac{z}{t}\right) d z}}} = {\color{red}{\int{\left(z + \frac{z^{2}}{t}\right)d z}}}$$

項別に積分せよ:

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

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

$$\int{\frac{z^{2}}{t} d z} + {\color{red}{\int{z d z}}}=\int{\frac{z^{2}}{t} d z} + {\color{red}{\frac{z^{1 + 1}}{1 + 1}}}=\int{\frac{z^{2}}{t} d z} + {\color{red}{\left(\frac{z^{2}}{2}\right)}}$$

定数倍の法則 $$$\int c f{\left(z \right)}\, dz = c \int f{\left(z \right)}\, dz$$$ を、$$$c=\frac{1}{t}$$$ と $$$f{\left(z \right)} = z^{2}$$$ に対して適用する:

$$\frac{z^{2}}{2} + {\color{red}{\int{\frac{z^{2}}{t} d z}}} = \frac{z^{2}}{2} + {\color{red}{\frac{\int{z^{2} d z}}{t}}}$$

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

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

$$$\int z \left(1 + \frac{z}{t}\right)\, dz = \frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t} + C$$$A