Integral of lnx-1 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 $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

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

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

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

         $$\int 1\, dx = x$$

      Now evaluate the sub-integral.

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

      $$\int 1\, dx = x$$

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

      $$\int \left(-1\right)\, dx = - x$$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is: