Integral of J(4x³+3x²-2x-8)dx dx

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

   $$\int i \left(\left(- 2 x + \left(4 x^{3} + 3 x^{2}\right)\right) - 8\right)\, dx = i \int \left(\left(- 2 x + \left(4 x^{3} + 3 x^{2}\right)\right) - 8\right)\, dx$$

   1. Integrate term-by-term:

      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 a constant times a function is the constant times the integral of the function:

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

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

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

               So, the result is: $x^{4}$

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

               $$\int 3 x^{2}\, dx = 3 \int x^{2}\, dx$$

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

                  $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

               So, the result is: $x^{3}$

            The result is: $x^{4} + x^{3}$

         The result is: $x^{4} + x^{3} - x^{2}$

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

         $$\int \left(-8\right)\, dx = - 8 x$$

      The result is: $x^{4} + x^{3} - x^{2} - 8 x$

   So, the result is: $i \left(x^{4} + x^{3} - x^{2} - 8 x\right)$

2. Now simplify:

   $$i x \left(x^{3} + x^{2} - x - 8\right)$$

3. Add the constant of integration:

   $$i x \left(x^{3} + x^{2} - x - 8\right)+ \mathrm{constant}$$

The answer is: