Integral of log(x) 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:

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

3. Now simplify:

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

4. Add the constant of integration:

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

The answer is:

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