Integral of $$$\ln\left(n\right)$$$

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

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

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

The integral can be rewritten as

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

Apply the constant rule $$$\int c\, dn = c n$$$ with $$$c=1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \ln\left(n\right)\, dn = n \left(\ln\left(n\right) - 1\right) + C$$$A