Integral of x^3+logx-2x dx

1. Integrate term-by-term:

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

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

      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: $- x^{2}$

   1. Integrate term-by-term:

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

         $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

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

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

2. Now simplify:

   $$x \left(\frac{x^{3}}{4} - x + \log{\left(x \right)} - 1\right)$$

3. Add the constant of integration:

   $$x \left(\frac{x^{3}}{4} - x + \log{\left(x \right)} - 1\right)+ \mathrm{constant}$$

The answer is: