Integral of -lgx dx

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

   $$\int \left(- \log{\left(x \right)}\right)\, dx = - \int \log{\left(x \right)}\, dx$$

   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:

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

   So, the result is: $- x \log{\left(x \right)} + x$

2. Now simplify:

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

3. Add the constant of integration:

   $$x \left(1 - \log{\left(x \right)}\right)+ \mathrm{constant}$$

The answer is: