Integral of $$$x e^{3 x}$$$

Find $$$\int x e^{3 x}\, dx$$$.

For the integral $$$\int{x e^{3 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=e^{3 x} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{3 x} d x}=\frac{e^{3 x}}{3}$$$ (steps can be seen »).

So,

$${\color{red}{\int{x e^{3 x} d x}}}={\color{red}{\left(x \cdot \frac{e^{3 x}}{3}-\int{\frac{e^{3 x}}{3} \cdot 1 d x}\right)}}={\color{red}{\left(\frac{x e^{3 x}}{3} - \int{\frac{e^{3 x}}{3} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(x \right)} = e^{3 x}$$$:

$$\frac{x e^{3 x}}{3} - {\color{red}{\int{\frac{e^{3 x}}{3} d x}}} = \frac{x e^{3 x}}{3} - {\color{red}{\left(\frac{\int{e^{3 x} d x}}{3}\right)}}$$

Let $$$u=3 x$$$.

Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

Therefore,

$$\frac{x e^{3 x}}{3} - \frac{{\color{red}{\int{e^{3 x} d x}}}}{3} = \frac{x e^{3 x}}{3} - \frac{{\color{red}{\int{\frac{e^{u}}{3} d u}}}}{3}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$\frac{x e^{3 x}}{3} - \frac{{\color{red}{\int{\frac{e^{u}}{3} d u}}}}{3} = \frac{x e^{3 x}}{3} - \frac{{\color{red}{\left(\frac{\int{e^{u} d u}}{3}\right)}}}{3}$$

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

$$\frac{x e^{3 x}}{3} - \frac{{\color{red}{\int{e^{u} d u}}}}{9} = \frac{x e^{3 x}}{3} - \frac{{\color{red}{e^{u}}}}{9}$$

Recall that $$$u=3 x$$$:

$$\frac{x e^{3 x}}{3} - \frac{e^{{\color{red}{u}}}}{9} = \frac{x e^{3 x}}{3} - \frac{e^{{\color{red}{\left(3 x\right)}}}}{9}$$

Therefore,

$$\int{x e^{3 x} d x} = \frac{x e^{3 x}}{3} - \frac{e^{3 x}}{9}$$

Simplify:

$$\int{x e^{3 x} d x} = \frac{\left(3 x - 1\right) e^{3 x}}{9}$$

Add the constant of integration:

$$\int{x e^{3 x} d x} = \frac{\left(3 x - 1\right) e^{3 x}}{9}+C$$

$$$\int x e^{3 x}\, dx = \frac{\left(3 x - 1\right) e^{3 x}}{9} + C$$$A