Integral of $$$e \ln\left(x\right)$$$

Find $$$\int e \ln\left(x\right)\, dx$$$.

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

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

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

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

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

The integral becomes

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int e \ln\left(x\right)\, dx = e x \left(\ln\left(x\right) - 1\right) + C$$$A