$$$\frac{1003605945944011425233769242881280649744658171441 t}{1000000000000000000000000000000000000000000000000}$$$ 的積分
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求$$$\int \frac{1003605945944011425233769242881280649744658171441 t}{1000000000000000000000000000000000000000000000000}\, dt$$$。
解答
套用常數倍法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$，使用 $$$c=\frac{1003605945944011425233769242881280649744658171441}{1000000000000000000000000000000000000000000000000}$$$ 與 $$$f{\left(t \right)} = t$$$：
$${\color{red}{\int{\frac{1003605945944011425233769242881280649744658171441 t}{1000000000000000000000000000000000000000000000000} d t}}} = {\color{red}{\left(\frac{1003605945944011425233769242881280649744658171441 \int{t d t}}{1000000000000000000000000000000000000000000000000}\right)}}$$
套用冪次法則 $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：
$$\frac{1003605945944011425233769242881280649744658171441 {\color{red}{\int{t d t}}}}{1000000000000000000000000000000000000000000000000}=\frac{1003605945944011425233769242881280649744658171441 {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}}{1000000000000000000000000000000000000000000000000}=\frac{1003605945944011425233769242881280649744658171441 {\color{red}{\left(\frac{t^{2}}{2}\right)}}}{1000000000000000000000000000000000000000000000000}$$
因此，
$$\int{\frac{1003605945944011425233769242881280649744658171441 t}{1000000000000000000000000000000000000000000000000} d t} = \frac{1003605945944011425233769242881280649744658171441 t^{2}}{2000000000000000000000000000000000000000000000000}$$
加上積分常數：
$$\int{\frac{1003605945944011425233769242881280649744658171441 t}{1000000000000000000000000000000000000000000000000} d t} = \frac{1003605945944011425233769242881280649744658171441 t^{2}}{2000000000000000000000000000000000000000000000000}+C$$
答案
$$$\int \frac{1003605945944011425233769242881280649744658171441 t}{1000000000000000000000000000000000000000000000000}\, dt = \frac{1003605945944011425233769242881280649744658171441 t^{2}}{2000000000000000000000000000000000000000000000000} + C$$$A