Integral of $$$- 9 x e^{- 3 x}$$$

Find $$$\int \left(- 9 x e^{- 3 x}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-9$$$ and $$$f{\left(x \right)} = x e^{- 3 x}$$$:

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

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 »).

The integral becomes

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

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 $$$f{\left(x \right)} = e^{- 3 x}$$$:

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

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

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

Thus,

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

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

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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