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

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

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

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

For the integral $$$\int{x e^{x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=e^{x} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{x} d x}=e^{x}$$$ (steps can be seen »).

So,

$$2 {\color{red}{\int{x e^{x} d x}}}=2 {\color{red}{\left(x \cdot e^{x}-\int{e^{x} \cdot 1 d x}\right)}}=2 {\color{red}{\left(x e^{x} - \int{e^{x} d x}\right)}}$$

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

$$2 x e^{x} - 2 {\color{red}{\int{e^{x} d x}}} = 2 x e^{x} - 2 {\color{red}{e^{x}}}$$

Therefore,

$$\int{2 x e^{x} d x} = 2 x e^{x} - 2 e^{x}$$

Simplify:

$$\int{2 x e^{x} d x} = 2 \left(x - 1\right) e^{x}$$

Add the constant of integration:

$$\int{2 x e^{x} d x} = 2 \left(x - 1\right) e^{x}+C$$

$$$\int 2 x e^{x}\, dx = 2 \left(x - 1\right) e^{x} + C$$$A