Integral of yln(y-1)dy dx

1. Use integration by parts:

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

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

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

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

   1. The integral of $y^{n}$ is $\frac{y^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int y\, dy = \frac{y^{2}}{2}$$

   Now evaluate the sub-integral.

2. The integral of a constant times a function is the constant times the integral of the function:

   $$\int \frac{y^{2}}{2 \left(y - 1\right)}\, dy = \frac{\int \frac{y^{2}}{y - 1}\, dy}{2}$$

   1. Rewrite the integrand:

      $$\frac{y^{2}}{y - 1} = y + 1 + \frac{1}{y - 1}$$

   2. Integrate term-by-term:

      1. The integral of $y^{n}$ is $\frac{y^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int y\, dy = \frac{y^{2}}{2}$$

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

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

      1. Let $u = y - 1$.

         Then let $du = dy$ and substitute $du$:

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(y - 1 \right)}$$

      The result is: $\frac{y^{2}}{2} + y + \log{\left(y - 1 \right)}$

   So, the result is: $\frac{y^{2}}{4} + \frac{y}{2} + \frac{\log{\left(y - 1 \right)}}{2}$

3. Now simplify:

   $$\frac{y^{2} \log{\left(y - 1 \right)}}{2} - \frac{y^{2}}{4} - \frac{y}{2} - \frac{\log{\left(y - 1 \right)}}{2}$$

4. Add the constant of integration:

   $$\frac{y^{2} \log{\left(y - 1 \right)}}{2} - \frac{y^{2}}{4} - \frac{y}{2} - \frac{\log{\left(y - 1 \right)}}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{y^{2} \log{\left(y - 1 \right)}}{2} - \frac{y^{2}}{4} - \frac{y}{2} - \frac{\log{\left(y - 1 \right)}}{2}+ \mathrm{constant}$$