Integral of lnx+x 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. 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:

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

   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:

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