Integral of $$$x^{2} e^{- 5 x}$$$

Find $$$\int x^{2} e^{- 5 x}\, dx$$$.

For the integral $$$\int{x^{2} e^{- 5 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{2}$$$ and $$$\operatorname{dv}=e^{- 5 x} dx$$$.

Then $$$\operatorname{du}=\left(x^{2}\right)^{\prime }dx=2 x dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- 5 x} d x}=- \frac{e^{- 5 x}}{5}$$$ (steps can be seen »).

The integral can be rewritten as

$${\color{red}{\int{x^{2} e^{- 5 x} d x}}}={\color{red}{\left(x^{2} \cdot \left(- \frac{e^{- 5 x}}{5}\right)-\int{\left(- \frac{e^{- 5 x}}{5}\right) \cdot 2 x d x}\right)}}={\color{red}{\left(- \frac{x^{2} e^{- 5 x}}{5} - \int{\left(- \frac{2 x e^{- 5 x}}{5}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{2}{5}$$$ and $$$f{\left(x \right)} = x e^{- 5 x}$$$:

$$- \frac{x^{2} e^{- 5 x}}{5} - {\color{red}{\int{\left(- \frac{2 x e^{- 5 x}}{5}\right)d x}}} = - \frac{x^{2} e^{- 5 x}}{5} - {\color{red}{\left(- \frac{2 \int{x e^{- 5 x} d x}}{5}\right)}}$$

For the integral $$$\int{x e^{- 5 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=e^{- 5 x} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- 5 x} d x}=- \frac{e^{- 5 x}}{5}$$$ (steps can be seen »).

So,

$$- \frac{x^{2} e^{- 5 x}}{5} + \frac{2 {\color{red}{\int{x e^{- 5 x} d x}}}}{5}=- \frac{x^{2} e^{- 5 x}}{5} + \frac{2 {\color{red}{\left(x \cdot \left(- \frac{e^{- 5 x}}{5}\right)-\int{\left(- \frac{e^{- 5 x}}{5}\right) \cdot 1 d x}\right)}}}{5}=- \frac{x^{2} e^{- 5 x}}{5} + \frac{2 {\color{red}{\left(- \frac{x e^{- 5 x}}{5} - \int{\left(- \frac{e^{- 5 x}}{5}\right)d x}\right)}}}{5}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{5}$$$ and $$$f{\left(x \right)} = e^{- 5 x}$$$:

$$- \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} - \frac{2 {\color{red}{\int{\left(- \frac{e^{- 5 x}}{5}\right)d x}}}}{5} = - \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} - \frac{2 {\color{red}{\left(- \frac{\int{e^{- 5 x} d x}}{5}\right)}}}{5}$$

Let $$$u=- 5 x$$$.

Then $$$du=\left(- 5 x\right)^{\prime }dx = - 5 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{5}$$$.

The integral can be rewritten as

$$- \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} + \frac{2 {\color{red}{\int{e^{- 5 x} d x}}}}{25} = - \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}}}{25}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{5}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$- \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}}}{25} = - \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} + \frac{2 {\color{red}{\left(- \frac{\int{e^{u} d u}}{5}\right)}}}{25}$$

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

$$- \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} - \frac{2 {\color{red}{\int{e^{u} d u}}}}{125} = - \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} - \frac{2 {\color{red}{e^{u}}}}{125}$$

Recall that $$$u=- 5 x$$$:

$$- \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} - \frac{2 e^{{\color{red}{u}}}}{125} = - \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} - \frac{2 e^{{\color{red}{\left(- 5 x\right)}}}}{125}$$

Therefore,

$$\int{x^{2} e^{- 5 x} d x} = - \frac{x^{2} e^{- 5 x}}{5} - \frac{2 x e^{- 5 x}}{25} - \frac{2 e^{- 5 x}}{125}$$

Simplify:

$$\int{x^{2} e^{- 5 x} d x} = \frac{\left(- 25 x^{2} - 10 x - 2\right) e^{- 5 x}}{125}$$

Add the constant of integration:

$$\int{x^{2} e^{- 5 x} d x} = \frac{\left(- 25 x^{2} - 10 x - 2\right) e^{- 5 x}}{125}+C$$

$$$\int x^{2} e^{- 5 x}\, dx = \frac{\left(- 25 x^{2} - 10 x - 2\right) e^{- 5 x}}{125} + C$$$A