Integral von $$$\frac{363 x}{e^{\frac{109}{1000}}}$$$

Bestimme $$$\int \frac{363 x}{e^{\frac{109}{1000}}}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{363}{e^{\frac{109}{1000}}}$$$ und $$$f{\left(x \right)} = x$$$ an:

$${\color{red}{\int{\frac{363 x}{e^{\frac{109}{1000}}} d x}}} = {\color{red}{\left(\frac{363 \int{x d x}}{e^{\frac{109}{1000}}}\right)}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=1$$$ an:

$$\frac{363 {\color{red}{\int{x d x}}}}{e^{\frac{109}{1000}}}=\frac{363 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{e^{\frac{109}{1000}}}=\frac{363 {\color{red}{\left(\frac{x^{2}}{2}\right)}}}{e^{\frac{109}{1000}}}$$

Daher,

$$\int{\frac{363 x}{e^{\frac{109}{1000}}} d x} = \frac{363 x^{2}}{2 e^{\frac{109}{1000}}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{363 x}{e^{\frac{109}{1000}}} d x} = \frac{363 x^{2}}{2 e^{\frac{109}{1000}}}+C$$

$$$\int \frac{363 x}{e^{\frac{109}{1000}}}\, dx = \frac{363 x^{2}}{2 e^{\frac{109}{1000}}} + C$$$A