Integral of $$$\ln\left(x_{0}\right)$$$

Find $$$\int \ln\left(x_{0}\right)\, dx_{0}$$$.

For the integral $$$\int{\ln{\left(x_{0} \right)} d x_{0}}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\ln{\left(x_{0} \right)}$$$ and $$$\operatorname{dv}=dx_{0}$$$.

Then $$$\operatorname{du}=\left(\ln{\left(x_{0} \right)}\right)^{\prime }dx_{0}=\frac{dx_{0}}{x_{0}}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d x_{0}}=x_{0}$$$ (steps can be seen »).

Therefore,

$${\color{red}{\int{\ln{\left(x_{0} \right)} d x_{0}}}}={\color{red}{\left(\ln{\left(x_{0} \right)} \cdot x_{0}-\int{x_{0} \cdot \frac{1}{x_{0}} d x_{0}}\right)}}={\color{red}{\left(x_{0} \ln{\left(x_{0} \right)} - \int{1 d x_{0}}\right)}}$$

Apply the constant rule $$$\int c\, dx_{0} = c x_{0}$$$ with $$$c=1$$$:

$$x_{0} \ln{\left(x_{0} \right)} - {\color{red}{\int{1 d x_{0}}}} = x_{0} \ln{\left(x_{0} \right)} - {\color{red}{x_{0}}}$$

Therefore,

$$\int{\ln{\left(x_{0} \right)} d x_{0}} = x_{0} \ln{\left(x_{0} \right)} - x_{0}$$

Simplify:

$$\int{\ln{\left(x_{0} \right)} d x_{0}} = x_{0} \left(\ln{\left(x_{0} \right)} - 1\right)$$

Add the constant of integration:

$$\int{\ln{\left(x_{0} \right)} d x_{0}} = x_{0} \left(\ln{\left(x_{0} \right)} - 1\right)+C$$

$$$\int \ln\left(x_{0}\right)\, dx_{0} = x_{0} \left(\ln\left(x_{0}\right) - 1\right) + C$$$A