Integral of xlnx^2 dx

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

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

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

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

      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 $u = 2 u$.

            Then let $du = 2 du$ and substitute $\frac{du}{2}$:

            $$\int \frac{e^{u}}{4}\, du$$

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

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

               1. The integral of the exponential function is itself.

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

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

            Now substitute $u$ back in:

            $$\frac{e^{2 u}}{2}$$

         Method #2

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

            Then let $du = 2 e^{2 u} du$ and substitute $\frac{du}{2}$:

            $$\int \frac{1}{4}\, du$$

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

               $$\int \frac{1}{2}\, du = \frac{\int 1\, du}{2}$$

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

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

               So, the result is: $\frac{u}{2}$

            Now substitute $u$ back in:

            $$\frac{e^{2 u}}{2}$$

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

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

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

      1. Let $u = 2 u$.

         Then let $du = 2 du$ and substitute $\frac{du}{2}$:

         $$\int \frac{e^{u}}{4}\, du$$

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

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

            1. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$\frac{e^{2 u}}{2}$$

      Now evaluate the sub-integral.

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

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

      1. Let $u = 2 u$.

         Then let $du = 2 du$ and substitute $\frac{du}{2}$:

         $$\int \frac{e^{u}}{4}\, du$$

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

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

            1. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$\frac{e^{2 u}}{2}$$

      So, the result is: $\frac{e^{2 u}}{4}$

   Now substitute $u$ back in:

   $$\frac{x^{2} \log{\left(x \right)}^{2}}{2} - \frac{x^{2} \log{\left(x \right)}}{2} + \frac{x^{2}}{4}$$

2. Now simplify:

   $$\frac{x^{2} \cdot \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x^{2} \cdot \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \cdot \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{4}+ \mathrm{constant}$$