$$$z \left(1 + \frac{z}{t}\right)$$$ 對 $$$z$$$ 的積分
您的輸入
求$$$\int z \left(1 + \frac{z}{t}\right)\, dz$$$。
解答
Expand the expression:
$${\color{red}{\int{z \left(1 + \frac{z}{t}\right) d z}}} = {\color{red}{\int{\left(z + \frac{z^{2}}{t}\right)d z}}}$$
逐項積分：
$${\color{red}{\int{\left(z + \frac{z^{2}}{t}\right)d z}}} = {\color{red}{\left(\int{z d z} + \int{\frac{z^{2}}{t} d z}\right)}}$$
套用冪次法則 $$$\int z^{n}\, dz = \frac{z^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：
$$\int{\frac{z^{2}}{t} d z} + {\color{red}{\int{z d z}}}=\int{\frac{z^{2}}{t} d z} + {\color{red}{\frac{z^{1 + 1}}{1 + 1}}}=\int{\frac{z^{2}}{t} d z} + {\color{red}{\left(\frac{z^{2}}{2}\right)}}$$
套用常數倍法則 $$$\int c f{\left(z \right)}\, dz = c \int f{\left(z \right)}\, dz$$$，使用 $$$c=\frac{1}{t}$$$ 與 $$$f{\left(z \right)} = z^{2}$$$：
$$\frac{z^{2}}{2} + {\color{red}{\int{\frac{z^{2}}{t} d z}}} = \frac{z^{2}}{2} + {\color{red}{\frac{\int{z^{2} d z}}{t}}}$$
套用冪次法則 $$$\int z^{n}\, dz = \frac{z^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：
$$\frac{z^{2}}{2} + \frac{{\color{red}{\int{z^{2} d z}}}}{t}=\frac{z^{2}}{2} + \frac{{\color{red}{\frac{z^{1 + 2}}{1 + 2}}}}{t}=\frac{z^{2}}{2} + \frac{{\color{red}{\left(\frac{z^{3}}{3}\right)}}}{t}$$
因此，
$$\int{z \left(1 + \frac{z}{t}\right) d z} = \frac{z^{2}}{2} + \frac{z^{3}}{3 t}$$
化簡：
$$\int{z \left(1 + \frac{z}{t}\right) d z} = \frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}$$
加上積分常數：
$$\int{z \left(1 + \frac{z}{t}\right) d z} = \frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t}+C$$
答案
$$$\int z \left(1 + \frac{z}{t}\right)\, dz = \frac{z^{2} \left(\frac{t}{2} + \frac{z}{3}\right)}{t} + C$$$A