Integral of (ln(y)-1)dy dx

1. Integrate term-by-term:

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(y \right)} = \log{\left(y \right)}$ and let $\operatorname{dv}{\left(y \right)} = 1$.

      Then $\operatorname{du}{\left(y \right)} = \frac{1}{y}$.

      To find $v{\left(y \right)}$:

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dy = y$$

      Now evaluate the sub-integral.

   2. The integral of a constant is the constant times the variable of integration:

      $$\int 1\, dy = y$$

   1. The integral of a constant is the constant times the variable of integration:

      $$\int \left(-1\right)\, dy = - y$$

   The result is: $y \log{\left(y \right)} - 2 y$

2. Now simplify:

   $$y \left(\log{\left(y \right)} - 2\right)$$

3. Add the constant of integration:

   $$y \left(\log{\left(y \right)} - 2\right)+ \mathrm{constant}$$

The answer is:

$$y \left(\log{\left(y \right)} - 2\right)+ \mathrm{constant}$$