$$$38 \left(\frac{6}{5}\right)^{t}$$$の積分

$$$\int 38 \left(\frac{6}{5}\right)^{t}\, dt$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ を、$$$c=38$$$ と $$$f{\left(t \right)} = \left(\frac{6}{5}\right)^{t}$$$ に対して適用する:

$${\color{red}{\int{38 \left(\frac{6}{5}\right)^{t} d t}}} = {\color{red}{\left(38 \int{\left(\frac{6}{5}\right)^{t} d t}\right)}}$$

Apply the exponential rule $$$\int{a^{t} d t} = \frac{a^{t}}{\ln{\left(a \right)}}$$$ with $$$a=\frac{6}{5}$$$:

$$38 {\color{red}{\int{\left(\frac{6}{5}\right)^{t} d t}}} = 38 {\color{red}{\frac{\left(\frac{6}{5}\right)^{t}}{\ln{\left(\frac{6}{5} \right)}}}}$$

したがって、

$$\int{38 \left(\frac{6}{5}\right)^{t} d t} = \frac{38 \left(\frac{6}{5}\right)^{t}}{\ln{\left(\frac{6}{5} \right)}}$$

簡単化せよ:

$$\int{38 \left(\frac{6}{5}\right)^{t} d t} = \frac{38 \left(\frac{6}{5}\right)^{t}}{- \ln{\left(5 \right)} + \ln{\left(6 \right)}}$$

積分定数を加える:

$$\int{38 \left(\frac{6}{5}\right)^{t} d t} = \frac{38 \left(\frac{6}{5}\right)^{t}}{- \ln{\left(5 \right)} + \ln{\left(6 \right)}}+C$$

$$$\int 38 \left(\frac{6}{5}\right)^{t}\, dt = \frac{38 \left(\frac{6}{5}\right)^{t}}{- \ln\left(5\right) + \ln\left(6\right)} + C$$$A