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

Find $$$\int \ln\left(y\right)\, dy$$$.

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

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

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

The integral becomes

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

Apply the constant rule $$$\int c\, dy = c y$$$ with $$$c=1$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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