Integral of $$$17 x^{3} e^{x}$$$

Find $$$\int 17 x^{3} e^{x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=17$$$ and $$$f{\left(x \right)} = x^{3} e^{x}$$$:

$${\color{red}{\int{17 x^{3} e^{x} d x}}} = {\color{red}{\left(17 \int{x^{3} e^{x} d x}\right)}}$$

For the integral $$$\int{x^{3} e^{x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{3}$$$ and $$$\operatorname{dv}=e^{x} dx$$$.

Then $$$\operatorname{du}=\left(x^{3}\right)^{\prime }dx=3 x^{2} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{x} d x}=e^{x}$$$ (steps can be seen »).

Thus,

$$17 {\color{red}{\int{x^{3} e^{x} d x}}}=17 {\color{red}{\left(x^{3} \cdot e^{x}-\int{e^{x} \cdot 3 x^{2} d x}\right)}}=17 {\color{red}{\left(x^{3} e^{x} - \int{3 x^{2} e^{x} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = x^{2} e^{x}$$$:

$$17 x^{3} e^{x} - 17 {\color{red}{\int{3 x^{2} e^{x} d x}}} = 17 x^{3} e^{x} - 17 {\color{red}{\left(3 \int{x^{2} e^{x} d x}\right)}}$$

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 »).

The integral becomes

$$17 x^{3} e^{x} - 51 {\color{red}{\int{x^{2} e^{x} d x}}}=17 x^{3} e^{x} - 51 {\color{red}{\left(x^{2} \cdot e^{x}-\int{e^{x} \cdot 2 x d x}\right)}}=17 x^{3} e^{x} - 51 {\color{red}{\left(x^{2} e^{x} - \int{2 x e^{x} d x}\right)}}$$

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

$$17 x^{3} e^{x} - 51 x^{2} e^{x} + 51 {\color{red}{\int{2 x e^{x} d x}}} = 17 x^{3} e^{x} - 51 x^{2} e^{x} + 51 {\color{red}{\left(2 \int{x e^{x} d x}\right)}}$$

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 »).

Thus,

$$17 x^{3} e^{x} - 51 x^{2} e^{x} + 102 {\color{red}{\int{x e^{x} d x}}}=17 x^{3} e^{x} - 51 x^{2} e^{x} + 102 {\color{red}{\left(x \cdot e^{x}-\int{e^{x} \cdot 1 d x}\right)}}=17 x^{3} e^{x} - 51 x^{2} e^{x} + 102 {\color{red}{\left(x e^{x} - \int{e^{x} d x}\right)}}$$

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

$$17 x^{3} e^{x} - 51 x^{2} e^{x} + 102 x e^{x} - 102 {\color{red}{\int{e^{x} d x}}} = 17 x^{3} e^{x} - 51 x^{2} e^{x} + 102 x e^{x} - 102 {\color{red}{e^{x}}}$$

Therefore,

$$\int{17 x^{3} e^{x} d x} = 17 x^{3} e^{x} - 51 x^{2} e^{x} + 102 x e^{x} - 102 e^{x}$$

Simplify:

$$\int{17 x^{3} e^{x} d x} = 17 \left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}$$

Add the constant of integration:

$$\int{17 x^{3} e^{x} d x} = 17 \left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}+C$$

$$$\int 17 x^{3} e^{x}\, dx = 17 \left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x} + C$$$A