Integral of $$$x^{3} e^{6 x}$$$

Find $$$\int x^{3} e^{6 x}\, dx$$$.

For the integral $$$\int{x^{3} e^{6 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{3}$$$ and $$$\operatorname{dv}=e^{6 x} dx$$$.

Then $$$\operatorname{du}=\left(x^{3}\right)^{\prime }dx=3 x^{2} dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{6 x} d x}=\frac{e^{6 x}}{6}$$$ (steps can be seen »).

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = x^{2} e^{6 x}$$$:

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

For the integral $$$\int{x^{2} e^{6 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{2}$$$ and $$$\operatorname{dv}=e^{6 x} dx$$$.

Then $$$\operatorname{du}=\left(x^{2}\right)^{\prime }dx=2 x dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{6 x} d x}=\frac{e^{6 x}}{6}$$$ (steps can be seen »).

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(x \right)} = x e^{6 x}$$$:

$$\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{{\color{red}{\int{\frac{x e^{6 x}}{3} d x}}}}{2} = \frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{{\color{red}{\left(\frac{\int{x e^{6 x} d x}}{3}\right)}}}{2}$$

For the integral $$$\int{x e^{6 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=e^{6 x} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{6 x} d x}=\frac{e^{6 x}}{6}$$$ (steps can be seen »).

So,

$$\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{{\color{red}{\int{x e^{6 x} d x}}}}{6}=\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{{\color{red}{\left(x \cdot \frac{e^{6 x}}{6}-\int{\frac{e^{6 x}}{6} \cdot 1 d x}\right)}}}{6}=\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{{\color{red}{\left(\frac{x e^{6 x}}{6} - \int{\frac{e^{6 x}}{6} d x}\right)}}}{6}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(x \right)} = e^{6 x}$$$:

$$\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{\int{\frac{e^{6 x}}{6} d x}}}}{6} = \frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{\left(\frac{\int{e^{6 x} d x}}{6}\right)}}}{6}$$

Let $$$u=6 x$$$.

Then $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{6}$$$.

Therefore,

$$\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{\int{e^{6 x} d x}}}}{36} = \frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{\int{\frac{e^{u}}{6} d u}}}}{36}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{\int{\frac{e^{u}}{6} d u}}}}{36} = \frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{\left(\frac{\int{e^{u} d u}}{6}\right)}}}{36}$$

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

$$\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{\int{e^{u} d u}}}}{216} = \frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{{\color{red}{e^{u}}}}{216}$$

Recall that $$$u=6 x$$$:

$$\frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{e^{{\color{red}{u}}}}{216} = \frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{e^{{\color{red}{\left(6 x\right)}}}}{216}$$

Therefore,

$$\int{x^{3} e^{6 x} d x} = \frac{x^{3} e^{6 x}}{6} - \frac{x^{2} e^{6 x}}{12} + \frac{x e^{6 x}}{36} - \frac{e^{6 x}}{216}$$

Simplify:

$$\int{x^{3} e^{6 x} d x} = \frac{\left(36 x^{3} - 18 x^{2} + 6 x - 1\right) e^{6 x}}{216}$$

Add the constant of integration:

$$\int{x^{3} e^{6 x} d x} = \frac{\left(36 x^{3} - 18 x^{2} + 6 x - 1\right) e^{6 x}}{216}+C$$

$$$\int x^{3} e^{6 x}\, dx = \frac{\left(36 x^{3} - 18 x^{2} + 6 x - 1\right) e^{6 x}}{216} + C$$$A