Integral of ln(x)*ln(x) dx

1. There are multiple ways to do this integral.

   Method #1

   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:

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

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = \log{\left(x \right)}^{2}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = \frac{2 \log{\left(x \right)}}{x}$.

      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. Use integration by parts:

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

      Let $u{\left(x \right)} = 2 \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

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

      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.

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

      $$\int 2\, dx = 2 x$$

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:

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