Integral of $$$\left(x^{2} - 128 x\right) e^{2 x}$$$

Find $$$\int \left(x^{2} - 128 x\right) e^{2 x}\, dx$$$.

For the integral $$$\int{\left(x^{2} - 128 x\right) e^{2 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x \left(x - 128\right)$$$ and $$$\operatorname{dv}=e^{2 x} dx$$$.

Then $$$\operatorname{du}=\left(x \left(x - 128\right)\right)^{\prime }dx=\left(2 x - 128\right) dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{2 x} d x}=\frac{e^{2 x}}{2}$$$ (steps can be seen »).

The integral becomes

$${\color{red}{\int{\left(x^{2} - 128 x\right) e^{2 x} d x}}}={\color{red}{\left(x \left(x - 128\right) \cdot \frac{e^{2 x}}{2}-\int{\frac{e^{2 x}}{2} \cdot \left(2 x - 128\right) d x}\right)}}={\color{red}{\left(\frac{x \left(x - 128\right) e^{2 x}}{2} - \int{\left(x - 64\right) e^{2 x} d x}\right)}}$$

For the integral $$$\int{\left(x - 64\right) e^{2 x} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x - 64$$$ and $$$\operatorname{dv}=e^{2 x} dx$$$.

Then $$$\operatorname{du}=\left(x - 64\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{2 x} d x}=\frac{e^{2 x}}{2}$$$ (steps can be seen »).

The integral becomes

$$\frac{x \left(x - 128\right) e^{2 x}}{2} - {\color{red}{\int{\left(x - 64\right) e^{2 x} d x}}}=\frac{x \left(x - 128\right) e^{2 x}}{2} - {\color{red}{\left(\left(x - 64\right) \cdot \frac{e^{2 x}}{2}-\int{\frac{e^{2 x}}{2} \cdot 1 d x}\right)}}=\frac{x \left(x - 128\right) e^{2 x}}{2} - {\color{red}{\left(\frac{\left(x - 64\right) e^{2 x}}{2} - \int{\frac{e^{2 x}}{2} d x}\right)}}$$

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 $$$f{\left(x \right)} = e^{2 x}$$$:

$$\frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + {\color{red}{\int{\frac{e^{2 x}}{2} d x}}} = \frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + {\color{red}{\left(\frac{\int{e^{2 x} d x}}{2}\right)}}$$

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

Thus,

$$\frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{{\color{red}{\int{e^{2 x} d x}}}}{2} = \frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{{\color{red}{\int{\frac{e^{u}}{2} d u}}}}{2}$$

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

$$\frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{{\color{red}{\int{\frac{e^{u}}{2} d u}}}}{2} = \frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{{\color{red}{\left(\frac{\int{e^{u} d u}}{2}\right)}}}{2}$$

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

$$\frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{{\color{red}{\int{e^{u} d u}}}}{4} = \frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{{\color{red}{e^{u}}}}{4}$$

Recall that $$$u=2 x$$$:

$$\frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{e^{{\color{red}{u}}}}{4} = \frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{e^{{\color{red}{\left(2 x\right)}}}}{4}$$

Therefore,

$$\int{\left(x^{2} - 128 x\right) e^{2 x} d x} = \frac{x \left(x - 128\right) e^{2 x}}{2} - \frac{\left(x - 64\right) e^{2 x}}{2} + \frac{e^{2 x}}{4}$$

Simplify:

$$\int{\left(x^{2} - 128 x\right) e^{2 x} d x} = \frac{\left(2 x^{2} - 258 x + 129\right) e^{2 x}}{4}$$

Add the constant of integration:

$$\int{\left(x^{2} - 128 x\right) e^{2 x} d x} = \frac{\left(2 x^{2} - 258 x + 129\right) e^{2 x}}{4}+C$$

$$$\int \left(x^{2} - 128 x\right) e^{2 x}\, dx = \frac{\left(2 x^{2} - 258 x + 129\right) e^{2 x}}{4} + C$$$A