Integral of log(logx) dx

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

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

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

   1. There are multiple ways to do this integral.

      Method #1

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

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

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

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

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

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

            1. Let $u = e^{u}$.

               Then let $du = e^{u} du$ and substitute $du$:

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

               1. The integral of the exponential function is itself.

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

               Now substitute $u$ back in:

               $$e^{e^{u}}$$

            Now evaluate the sub-integral.

         2. Let $u = e^{u}$.

            Then let $du = e^{u} du$ and substitute $du$:

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

            EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)

            Now substitute $u$ back in:

            $$\operatorname{Ei}{\left(e^{u} \right)}$$

         Now substitute $u$ back in:

         $$e^{u} \log{\left(u \right)} - \operatorname{Ei}{\left(u \right)}$$

      Method #2

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

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

         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.

         EiRule(a=1, b=0, context=exp(_u)/_u, symbol=_u)

   Now substitute $u$ back in:

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

2. Add the constant of integration:

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

The answer is:

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