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

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

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

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

The integral can be rewritten as

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \ln\left(u\right)\, du = u \left(\ln\left(u\right) - 1\right) + C$$$A