Integral of ln12x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 12 x$.

      Then let $du = 12 dx$ and substitute $\frac{du}{12}$:

      $$\int \frac{\log{\left(u \right)}}{12}\, du$$

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

         $$\int \log{\left(u \right)}\, du = \frac{\int \log{\left(u \right)}\, du}{12}$$

         1. Use integration by parts:

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

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

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

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

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

               $$\int 1\, du = u$$

            Now evaluate the sub-integral.

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

            $$\int 1\, du = u$$

         So, the result is: $\frac{u \log{\left(u \right)}}{12} - \frac{u}{12}$

      Now substitute $u$ back in:

      $$x \log{\left(12 x \right)} - x$$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = \log{\left(12 x \right)}$ 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:

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

2. Now simplify:

   $$x \left(\log{\left(12 x \right)} - 1\right)$$

3. Add the constant of integration:

   $$x \left(\log{\left(12 x \right)} - 1\right)+ \mathrm{constant}$$

The answer is:

$$x \left(\log{\left(12 x \right)} - 1\right)+ \mathrm{constant}$$