Integral of $$$\frac{x e^{\frac{1}{20}}}{6}$$$

Find $$$\int \frac{x e^{\frac{1}{20}}}{6}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{e^{\frac{1}{20}}}{6}$$$ and $$$f{\left(x \right)} = x$$$:

$${\color{red}{\int{\frac{x e^{\frac{1}{20}}}{6} d x}}} = {\color{red}{\left(\frac{e^{\frac{1}{20}} \int{x d x}}{6}\right)}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\frac{e^{\frac{1}{20}} {\color{red}{\int{x d x}}}}{6}=\frac{e^{\frac{1}{20}} {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{6}=\frac{e^{\frac{1}{20}} {\color{red}{\left(\frac{x^{2}}{2}\right)}}}{6}$$

Therefore,

$$\int{\frac{x e^{\frac{1}{20}}}{6} d x} = \frac{x^{2} e^{\frac{1}{20}}}{12}$$

Add the constant of integration:

$$\int{\frac{x e^{\frac{1}{20}}}{6} d x} = \frac{x^{2} e^{\frac{1}{20}}}{12}+C$$

$$$\int \frac{x e^{\frac{1}{20}}}{6}\, dx = \frac{x^{2} e^{\frac{1}{20}}}{12} + C$$$A