Integral of $$$8 \ln\left(e^{2}\right) - \ln\left(e^{12}\right)$$$

Find $$$\int \left(8 \ln\left(e^{2}\right) - \ln\left(e^{12}\right)\right)\, de$$$.

The input is rewritten: $$$\int{\left(8 \ln{\left(e^{2} \right)} - \ln{\left(e^{12} \right)}\right)d e}=\int{4 \ln{\left(e \right)} d e}$$$.

Apply the constant multiple rule $$$\int c f{\left(e \right)}\, de = c \int f{\left(e \right)}\, de$$$ with $$$c=4$$$ and $$$f{\left(e \right)} = \ln{\left(e \right)}$$$:

$${\color{red}{\int{4 \ln{\left(e \right)} d e}}} = {\color{red}{\left(4 \int{\ln{\left(e \right)} d e}\right)}}$$

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

Let $$$\operatorname{u}=\ln{\left(e \right)}$$$ and $$$\operatorname{dv}=de$$$.

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

The integral can be rewritten as

$$4 {\color{red}{\int{\ln{\left(e \right)} d e}}}=4 {\color{red}{\left(\ln{\left(e \right)} \cdot e-\int{e \cdot \frac{1}{e} d e}\right)}}=4 {\color{red}{\left(e \ln{\left(e \right)} - \int{1 d e}\right)}}$$

Apply the constant rule $$$\int c\, de = c e$$$ with $$$c=1$$$:

$$4 e \ln{\left(e \right)} - 4 {\color{red}{\int{1 d e}}} = 4 e \ln{\left(e \right)} - 4 {\color{red}{e}}$$

Therefore,

$$\int{4 \ln{\left(e \right)} d e} = 4 e \ln{\left(e \right)} - 4 e$$

Simplify:

$$\int{4 \ln{\left(e \right)} d e} = 4 e \left(\ln{\left(e \right)} - 1\right)$$

Add the constant of integration:

$$\int{4 \ln{\left(e \right)} d e} = 4 e \left(\ln{\left(e \right)} - 1\right)+C$$

$$$\int \left(8 \ln\left(e^{2}\right) - \ln\left(e^{12}\right)\right)\, de = 4 e \left(\ln\left(e\right) - 1\right) + C$$$A