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

Find $$$\int z^{2} e^{- z}\, dz$$$.

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

Let $$$\operatorname{u}=z^{2}$$$ and $$$\operatorname{dv}=e^{- z} dz$$$.

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

Therefore,

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

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

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

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

Let $$$\operatorname{u}=z$$$ and $$$\operatorname{dv}=e^{- z} dz$$$.

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

Therefore,

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

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

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

Let $$$u=- z$$$.

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

Therefore,

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

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

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

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

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

Recall that $$$u=- z$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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