Integral of $$$2 e^{7 x}$$$

Find $$$\int 2 e^{7 x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = e^{7 x}$$$:

$${\color{red}{\int{2 e^{7 x} d x}}} = {\color{red}{\left(2 \int{e^{7 x} d x}\right)}}$$

Let $$$u=7 x$$$.

Then $$$du=\left(7 x\right)^{\prime }dx = 7 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{7}$$$.

Thus,

$$2 {\color{red}{\int{e^{7 x} d x}}} = 2 {\color{red}{\int{\frac{e^{u}}{7} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{7}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$2 {\color{red}{\int{\frac{e^{u}}{7} d u}}} = 2 {\color{red}{\left(\frac{\int{e^{u} d u}}{7}\right)}}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$$\frac{2 {\color{red}{\int{e^{u} d u}}}}{7} = \frac{2 {\color{red}{e^{u}}}}{7}$$

Recall that $$$u=7 x$$$:

$$\frac{2 e^{{\color{red}{u}}}}{7} = \frac{2 e^{{\color{red}{\left(7 x\right)}}}}{7}$$

Therefore,

$$\int{2 e^{7 x} d x} = \frac{2 e^{7 x}}{7}$$

Add the constant of integration:

$$\int{2 e^{7 x} d x} = \frac{2 e^{7 x}}{7}+C$$

$$$\int 2 e^{7 x}\, dx = \frac{2 e^{7 x}}{7} + C$$$A