Integral of $$$f^{2} x^{2} e^{x}$$$ with respect to $$$x$$$

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

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

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

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 »).

Thus,

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

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

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

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,

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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