Integral of $$$49 x^{3} e^{- 7 x}$$$

Find $$$\int 49 x^{3} e^{- 7 x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=49$$$ and $$$f{\left(x \right)} = x^{3} e^{- 7 x}$$$:

$${\color{red}{\int{49 x^{3} e^{- 7 x} d x}}} = {\color{red}{\left(49 \int{x^{3} e^{- 7 x} d x}\right)}}$$

For the integral $$$\int{x^{3} e^{- 7 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{3}$$$ and $$$\operatorname{dv}=e^{- 7 x} dx$$$.

Then $$$\operatorname{du}=\left(x^{3}\right)^{\prime }dx=3 x^{2} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- 7 x} d x}=- \frac{e^{- 7 x}}{7}$$$ (steps can be seen »).

Therefore,

$$49 {\color{red}{\int{x^{3} e^{- 7 x} d x}}}=49 {\color{red}{\left(x^{3} \cdot \left(- \frac{e^{- 7 x}}{7}\right)-\int{\left(- \frac{e^{- 7 x}}{7}\right) \cdot 3 x^{2} d x}\right)}}=49 {\color{red}{\left(- \frac{x^{3} e^{- 7 x}}{7} - \int{\left(- \frac{3 x^{2} e^{- 7 x}}{7}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{3}{7}$$$ and $$$f{\left(x \right)} = x^{2} e^{- 7 x}$$$:

$$- 7 x^{3} e^{- 7 x} - 49 {\color{red}{\int{\left(- \frac{3 x^{2} e^{- 7 x}}{7}\right)d x}}} = - 7 x^{3} e^{- 7 x} - 49 {\color{red}{\left(- \frac{3 \int{x^{2} e^{- 7 x} d x}}{7}\right)}}$$

For the integral $$$\int{x^{2} e^{- 7 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{2}$$$ and $$$\operatorname{dv}=e^{- 7 x} dx$$$.

Then $$$\operatorname{du}=\left(x^{2}\right)^{\prime }dx=2 x dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- 7 x} d x}=- \frac{e^{- 7 x}}{7}$$$ (steps can be seen »).

The integral becomes

$$- 7 x^{3} e^{- 7 x} + 21 {\color{red}{\int{x^{2} e^{- 7 x} d x}}}=- 7 x^{3} e^{- 7 x} + 21 {\color{red}{\left(x^{2} \cdot \left(- \frac{e^{- 7 x}}{7}\right)-\int{\left(- \frac{e^{- 7 x}}{7}\right) \cdot 2 x d x}\right)}}=- 7 x^{3} e^{- 7 x} + 21 {\color{red}{\left(- \frac{x^{2} e^{- 7 x}}{7} - \int{\left(- \frac{2 x e^{- 7 x}}{7}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{2}{7}$$$ and $$$f{\left(x \right)} = x e^{- 7 x}$$$:

$$- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - 21 {\color{red}{\int{\left(- \frac{2 x e^{- 7 x}}{7}\right)d x}}} = - 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - 21 {\color{red}{\left(- \frac{2 \int{x e^{- 7 x} d x}}{7}\right)}}$$

For the integral $$$\int{x e^{- 7 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=e^{- 7 x} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{- 7 x} d x}=- \frac{e^{- 7 x}}{7}$$$ (steps can be seen »).

Therefore,

$$- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} + 6 {\color{red}{\int{x e^{- 7 x} d x}}}=- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} + 6 {\color{red}{\left(x \cdot \left(- \frac{e^{- 7 x}}{7}\right)-\int{\left(- \frac{e^{- 7 x}}{7}\right) \cdot 1 d x}\right)}}=- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} + 6 {\color{red}{\left(- \frac{x e^{- 7 x}}{7} - \int{\left(- \frac{e^{- 7 x}}{7}\right)d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{1}{7}$$$ and $$$f{\left(x \right)} = e^{- 7 x}$$$:

$$- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} - 6 {\color{red}{\int{\left(- \frac{e^{- 7 x}}{7}\right)d x}}} = - 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} - 6 {\color{red}{\left(- \frac{\int{e^{- 7 x} d x}}{7}\right)}}$$

Let $$$u=- 7 x$$$.

Then $$$du=\left(- 7 x\right)^{\prime }dx = - 7 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{7}$$$.

So,

$$- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} + \frac{6 {\color{red}{\int{e^{- 7 x} d x}}}}{7} = - 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} + \frac{6 {\color{red}{\int{\left(- \frac{e^{u}}{7}\right)d u}}}}{7}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{7}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} + \frac{6 {\color{red}{\int{\left(- \frac{e^{u}}{7}\right)d u}}}}{7} = - 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} + \frac{6 {\color{red}{\left(- \frac{\int{e^{u} d u}}{7}\right)}}}{7}$$

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

$$- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} - \frac{6 {\color{red}{\int{e^{u} d u}}}}{49} = - 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} - \frac{6 {\color{red}{e^{u}}}}{49}$$

Recall that $$$u=- 7 x$$$:

$$- 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} - \frac{6 e^{{\color{red}{u}}}}{49} = - 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} - \frac{6 e^{{\color{red}{\left(- 7 x\right)}}}}{49}$$

Therefore,

$$\int{49 x^{3} e^{- 7 x} d x} = - 7 x^{3} e^{- 7 x} - 3 x^{2} e^{- 7 x} - \frac{6 x e^{- 7 x}}{7} - \frac{6 e^{- 7 x}}{49}$$

Simplify:

$$\int{49 x^{3} e^{- 7 x} d x} = \frac{\left(- 343 x^{3} - 147 x^{2} - 42 x - 6\right) e^{- 7 x}}{49}$$

Add the constant of integration:

$$\int{49 x^{3} e^{- 7 x} d x} = \frac{\left(- 343 x^{3} - 147 x^{2} - 42 x - 6\right) e^{- 7 x}}{49}+C$$

$$$\int 49 x^{3} e^{- 7 x}\, dx = \frac{\left(- 343 x^{3} - 147 x^{2} - 42 x - 6\right) e^{- 7 x}}{49} + C$$$A