Integral of x^2*log^2x dx

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

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

   $$\int u^{2} e^{3 u}\, du$$

   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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$.

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

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

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = 3 u$.

            Then let $du = 3 du$ and substitute $\frac{du}{3}$:

            $$\int \frac{e^{u}}{9}\, du$$

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

               $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

               1. The integral of the exponential function is itself.

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

               So, the result is: $\frac{e^{u}}{3}$

            Now substitute $u$ back in:

            $$\frac{e^{3 u}}{3}$$

         Method #2

         1. Let $u = e^{3 u}$.

            Then let $du = 3 e^{3 u} du$ and substitute $\frac{du}{3}$:

            $$\int \frac{1}{9}\, du$$

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

               $$\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$$

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

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

               So, the result is: $\frac{u}{3}$

            Now substitute $u$ back in:

            $$\frac{e^{3 u}}{3}$$

      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 $u{\left(u \right)} = \frac{2 u}{3}$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$.

      Then $\operatorname{du}{\left(u \right)} = \frac{2}{3}$.

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

      1. Let $u = 3 u$.

         Then let $du = 3 du$ and substitute $\frac{du}{3}$:

         $$\int \frac{e^{u}}{9}\, du$$

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

            $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

            1. The integral of the exponential function is itself.

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

            So, the result is: $\frac{e^{u}}{3}$

         Now substitute $u$ back in:

         $$\frac{e^{3 u}}{3}$$

      Now evaluate the sub-integral.

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

      $$\int \frac{2 e^{3 u}}{9}\, du = \frac{2 \int e^{3 u}\, du}{9}$$

      1. Let $u = 3 u$.

         Then let $du = 3 du$ and substitute $\frac{du}{3}$:

         $$\int \frac{e^{u}}{9}\, du$$

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

            $$\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$$

            1. The integral of the exponential function is itself.

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

            So, the result is: $\frac{e^{u}}{3}$

         Now substitute $u$ back in:

         $$\frac{e^{3 u}}{3}$$

      So, the result is: $\frac{2 e^{3 u}}{27}$

   Now substitute $u$ back in:

   $$\frac{x^{3} \log{\left(x \right)}^{2}}{3} - \frac{2 x^{3} \log{\left(x \right)}}{9} + \frac{2 x^{3}}{27}$$

2. Now simplify:

   $$\frac{x^{3} \cdot \left(9 \log{\left(x \right)}^{2} - 6 \log{\left(x \right)} + 2\right)}{27}$$

3. Add the constant of integration:

   $$\frac{x^{3} \cdot \left(9 \log{\left(x \right)}^{2} - 6 \log{\left(x \right)} + 2\right)}{27}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{3} \cdot \left(9 \log{\left(x \right)}^{2} - 6 \log{\left(x \right)} + 2\right)}{27}+ \mathrm{constant}$$