Integral of dx/x(ln^2)x dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \left(- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u^{2}}\right)\, du$$

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

         $$\int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u^{2}}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u^{2}}\, du$$

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

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

            $$\int \left(- u^{2} e^{u}\right)\, du$$

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

               $$\int u^{2} e^{u}\, du = - \int u^{2} e^{u}\, du$$

               1. Use integration by parts:

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

                  Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

                  Then $\operatorname{du}{\left(u \right)} = 2 u$.

                  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.

               2. Use integration by parts:

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

                  Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

                  Then $\operatorname{du}{\left(u \right)} = 2$.

                  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.

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

                  $$\int 2 e^{u}\, du = 2 \int e^{u}\, du$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $2 e^{u}$

               So, the result is: $- u^{2} e^{u} + 2 u e^{u} - 2 e^{u}$

            Now substitute $u$ back in:

            $$- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{u} + \frac{2 \log{\left(\frac{1}{u} \right)}}{u} - \frac{2}{u}$$

         So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{2}}{u} - \frac{2 \log{\left(\frac{1}{u} \right)}}{u} + \frac{2}{u}$

      Now substitute $u$ back in:

      $$x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$$

   Method #2

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

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

      $$\int u^{2} e^{u}\, du$$

      1. Use integration by parts:

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

         Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

         Then $\operatorname{du}{\left(u \right)} = 2 u$.

         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.

      2. Use integration by parts:

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

         Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$.

         Then $\operatorname{du}{\left(u \right)} = 2$.

         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.

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

         $$\int 2 e^{u}\, du = 2 \int e^{u}\, du$$

         1. The integral of the exponential function is itself.

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

         So, the result is: $2 e^{u}$

      Now substitute $u$ back in:

      $$x \log{\left(x \right)}^{2} - 2 x \log{\left(x \right)} + 2 x$$

2. Now simplify:

   $$x \left(\log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 2\right)$$

3. Add the constant of integration:

   $$x \left(\log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 2\right)+ \mathrm{constant}$$

The answer is:

$$x \left(\log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 2\right)+ \mathrm{constant}$$