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,

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

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

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

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,

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

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

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

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