Integral of x-lnx 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. The integral of a constant times a function is the constant times the integral of the function:

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

      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 $\operatorname{dv}{\left(x \right)} = 1$.

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

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

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

            $$\int 1\, dx = x$$

         Now evaluate the sub-integral.

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

         $$\int 1\, dx = x$$

      So, the result is: $- x \log{\left(x \right)} + x$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is: