$$$\frac{_e a^{2} l t}{\nu}$$$ 关于$$$t$$$的积分
您的输入
求$$$\int \frac{_e a^{2} l t}{\nu}\, dt$$$。
解答
对 $$$c=\frac{_e a^{2} l}{\nu}$$$ 和 $$$f{\left(t \right)} = t$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：
$${\color{red}{\int{\frac{_e a^{2} l t}{\nu} d t}}} = {\color{red}{\frac{_e a^{2} l \int{t d t}}{\nu}}}$$
应用幂法则 $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：
$$\frac{_e a^{2} l {\color{red}{\int{t d t}}}}{\nu}=\frac{_e a^{2} l {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}}{\nu}=\frac{_e a^{2} l {\color{red}{\left(\frac{t^{2}}{2}\right)}}}{\nu}$$
因此，
$$\int{\frac{_e a^{2} l t}{\nu} d t} = \frac{_e a^{2} l t^{2}}{2 \nu}$$
加上积分常数：
$$\int{\frac{_e a^{2} l t}{\nu} d t} = \frac{_e a^{2} l t^{2}}{2 \nu}+C$$
答案
$$$\int \frac{_e a^{2} l t}{\nu}\, dt = \frac{_e a^{2} l t^{2}}{2 \nu} + C$$$A