Integral of x+2xlnx dx

1. Integrate term-by-term:

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

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

   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 $2 du$:

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

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

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

            1. Use integration by parts:

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

               Let $u{\left(u \right)} = u$ 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 $\frac{du}{2}$:

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

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

                     $$\text{False}$$

                     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.

            2. 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 $\frac{du}{2}$:

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

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

                     $$\text{False}$$

                     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: $\frac{e^{2 u}}{4}$

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

         Now substitute $u$ back in:

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

      Method #2

      1. Use integration by parts:

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

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

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

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

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

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

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

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

            So, the result is: $x^{2}$

         Now evaluate the sub-integral.

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

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

   The result is: $x^{2} \log{\left(x \right)}$

2. Add the constant of integration:

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

The answer is:

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