Integral of -xlogx 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 $- du$:

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

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

         $$\int \left(- u e^{2 u}\right)\, du = - \int u e^{2 u}\, du$$

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

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

      Now substitute $u$ back in:

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

   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 $\operatorname{dv}{\left(x \right)} = - x$.

      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:

         $$\int \left(- x\right)\, dx = - \int x\, dx$$

         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: $- \frac{x^{2}}{2}$

      Now evaluate the sub-integral.

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

      $$\int \left(- \frac{x}{2}\right)\, dx = - \frac{\int x\, dx}{2}$$

      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: $- \frac{x^{2}}{4}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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