Integral of ln(ln(x))/x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \log{\left(\log{\left(x \right)} \right)}$.

      Then let $du = \frac{dx}{x \log{\left(x \right)}}$ and substitute $du$:

      $$\int u e^{u}\, du$$

      1. Use integration by parts:

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

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

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

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

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         Now evaluate the sub-integral.

      2. The integral of the exponential function is itself.

         $$\int e^{u}\, du = e^{u}$$

      Now substitute $u$ back in:

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

   Method #2

   1. Let $u = \log{\left(x \right)}$.

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

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

      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:

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

      Now substitute $u$ back in:

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

   Method #3

   1. Let $u = \frac{1}{x}$.

      Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$:

      $$\int \left(- \frac{\log{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\right)\, du$$

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

         $$\int \frac{\log{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\, du = - \int \frac{\log{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\, du$$

         1. Let $u = \log{\left(\log{\left(\frac{1}{u} \right)} \right)}$.

            Then let $du = - \frac{du}{u \log{\left(\frac{1}{u} \right)}}$ and substitute $- du$:

            $$\int \left(- u e^{u}\right)\, du$$

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

               $$\int u e^{u}\, du = - \int u e^{u}\, du$$

               1. Use integration by parts:

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

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

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

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

                  1. The integral of the exponential function is itself.

                     $$\int e^{u}\, du = e^{u}$$

                  Now evaluate the sub-integral.

               2. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

               So, the result is: $- u e^{u} + e^{u}$

            Now substitute $u$ back in:

            $$- \log{\left(\frac{1}{u} \right)} \log{\left(\log{\left(\frac{1}{u} \right)} \right)} + \log{\left(\frac{1}{u} \right)}$$

         So, the result is: $\log{\left(\frac{1}{u} \right)} \log{\left(\log{\left(\frac{1}{u} \right)} \right)} - \log{\left(\frac{1}{u} \right)}$

      Now substitute $u$ back in:

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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