Integral of $$$\frac{363 x}{e^{\frac{109}{1000}}}$$$
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Find $$$\int \frac{363 x}{e^{\frac{109}{1000}}}\, dx$$$.
Solution
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{363}{e^{\frac{109}{1000}}}$$$ and $$$f{\left(x \right)} = x$$$:
$${\color{red}{\int{\frac{363 x}{e^{\frac{109}{1000}}} d x}}} = {\color{red}{\left(\frac{363 \int{x d x}}{e^{\frac{109}{1000}}}\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$\frac{363 {\color{red}{\int{x d x}}}}{e^{\frac{109}{1000}}}=\frac{363 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{e^{\frac{109}{1000}}}=\frac{363 {\color{red}{\left(\frac{x^{2}}{2}\right)}}}{e^{\frac{109}{1000}}}$$
Therefore,
$$\int{\frac{363 x}{e^{\frac{109}{1000}}} d x} = \frac{363 x^{2}}{2 e^{\frac{109}{1000}}}$$
Add the constant of integration:
$$\int{\frac{363 x}{e^{\frac{109}{1000}}} d x} = \frac{363 x^{2}}{2 e^{\frac{109}{1000}}}+C$$
Answer
$$$\int \frac{363 x}{e^{\frac{109}{1000}}}\, dx = \frac{363 x^{2}}{2 e^{\frac{109}{1000}}} + C$$$A