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

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

For the integral $$$\int{x^{2} e^{- x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{2}$$$ and $$$\operatorname{dv}=e^{- x} dx$$$.

Then $$$\operatorname{du}=\left(x^{2}\right)^{\prime }dx=2 x dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- x} d x}=- e^{- x}$$$ (steps can be seen »).

Therefore,

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

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

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

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 »).

Therefore,

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

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

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

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

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

The integral becomes

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

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

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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