Integral of xln(x^2) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

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

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

         1. Use integration by parts:

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

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

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

            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. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

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

   Method #2

   1. Use integration by parts:

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

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

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

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

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int x\, dx = \frac{x^{2}}{2}$$

      Now evaluate the sub-integral.

   2. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int x\, dx = \frac{x^{2}}{2}$$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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