$$$\frac{_e a^{2} l t}{\nu}$$$ 對 $$$t$$$ 的積分

求$$$\int \frac{_e a^{2} l t}{\nu}\, dt$$$。

套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$，使用 $$$c=\frac{_e a^{2} l}{\nu}$$$ 與 $$$f{\left(t \right)} = t$$$：

$${\color{red}{\int{\frac{_e a^{2} l t}{\nu} d t}}} = {\color{red}{\frac{_e a^{2} l \int{t d t}}{\nu}}}$$

套用冪次法則 $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：

$$\frac{_e a^{2} l {\color{red}{\int{t d t}}}}{\nu}=\frac{_e a^{2} l {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}}{\nu}=\frac{_e a^{2} l {\color{red}{\left(\frac{t^{2}}{2}\right)}}}{\nu}$$

因此，

$$\int{\frac{_e a^{2} l t}{\nu} d t} = \frac{_e a^{2} l t^{2}}{2 \nu}$$

加上積分常數：

$$\int{\frac{_e a^{2} l t}{\nu} d t} = \frac{_e a^{2} l t^{2}}{2 \nu}+C$$

$$$\int \frac{_e a^{2} l t}{\nu}\, dt = \frac{_e a^{2} l t^{2}}{2 \nu} + C$$$A