$$$\left(x + 3\right) \ln\left(5\right)$$$の積分

$$$\int \left(x + 3\right) \ln\left(5\right)\, dx$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\ln{\left(5 \right)}$$$ と $$$f{\left(x \right)} = x + 3$$$ に対して適用する:

$${\color{red}{\int{\left(x + 3\right) \ln{\left(5 \right)} d x}}} = {\color{red}{\ln{\left(5 \right)} \int{\left(x + 3\right)d x}}}$$

項別に積分せよ:

$$\ln{\left(5 \right)} {\color{red}{\int{\left(x + 3\right)d x}}} = \ln{\left(5 \right)} {\color{red}{\left(\int{3 d x} + \int{x d x}\right)}}$$

$$$c=3$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$\ln{\left(5 \right)} \left(\int{x d x} + {\color{red}{\int{3 d x}}}\right) = \ln{\left(5 \right)} \left(\int{x d x} + {\color{red}{\left(3 x\right)}}\right)$$

$$$n=1$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$$\ln{\left(5 \right)} \left(3 x + {\color{red}{\int{x d x}}}\right)=\ln{\left(5 \right)} \left(3 x + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}\right)=\ln{\left(5 \right)} \left(3 x + {\color{red}{\left(\frac{x^{2}}{2}\right)}}\right)$$

したがって、

$$\int{\left(x + 3\right) \ln{\left(5 \right)} d x} = \left(\frac{x^{2}}{2} + 3 x\right) \ln{\left(5 \right)}$$

簡単化せよ:

$$\int{\left(x + 3\right) \ln{\left(5 \right)} d x} = \frac{x \left(x + 6\right) \ln{\left(5 \right)}}{2}$$

積分定数を加える:

$$\int{\left(x + 3\right) \ln{\left(5 \right)} d x} = \frac{x \left(x + 6\right) \ln{\left(5 \right)}}{2}+C$$

$$$\int \left(x + 3\right) \ln\left(5\right)\, dx = \frac{x \left(x + 6\right) \ln\left(5\right)}{2} + C$$$A