$$$38 \left(\frac{6}{5}\right)^{t}$$$ 的積分
您的輸入
求$$$\int 38 \left(\frac{6}{5}\right)^{t}\, dt$$$。
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 $$$c=38$$$ 與 $$$f{\left(t \right)} = \left(\frac{6}{5}\right)^{t}$$$:
$${\color{red}{\int{38 \left(\frac{6}{5}\right)^{t} d t}}} = {\color{red}{\left(38 \int{\left(\frac{6}{5}\right)^{t} d t}\right)}}$$
Apply the exponential rule $$$\int{a^{t} d t} = \frac{a^{t}}{\ln{\left(a \right)}}$$$ with $$$a=\frac{6}{5}$$$:
$$38 {\color{red}{\int{\left(\frac{6}{5}\right)^{t} d t}}} = 38 {\color{red}{\frac{\left(\frac{6}{5}\right)^{t}}{\ln{\left(\frac{6}{5} \right)}}}}$$
因此,
$$\int{38 \left(\frac{6}{5}\right)^{t} d t} = \frac{38 \left(\frac{6}{5}\right)^{t}}{\ln{\left(\frac{6}{5} \right)}}$$
化簡:
$$\int{38 \left(\frac{6}{5}\right)^{t} d t} = \frac{38 \left(\frac{6}{5}\right)^{t}}{- \ln{\left(5 \right)} + \ln{\left(6 \right)}}$$
加上積分常數:
$$\int{38 \left(\frac{6}{5}\right)^{t} d t} = \frac{38 \left(\frac{6}{5}\right)^{t}}{- \ln{\left(5 \right)} + \ln{\left(6 \right)}}+C$$
答案
$$$\int 38 \left(\frac{6}{5}\right)^{t}\, dt = \frac{38 \left(\frac{6}{5}\right)^{t}}{- \ln\left(5\right) + \ln\left(6\right)} + C$$$A