Integral of x(x^2+3) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = x^{2}$.

      Then let $du = 2 x dx$ and substitute $du$:

      $$\int \left(\frac{u}{2} + \frac{3}{2}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int \frac{u}{2}\, du = \frac{\int u\, du}{2}$$

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

               $$\int u\, du = \frac{u^{2}}{2}$$

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

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

            $$\int \frac{3}{2}\, du = \frac{3 u}{2}$$

         The result is: $\frac{u^{2}}{4} + \frac{3 u}{2}$

      Now substitute $u$ back in:

      $$\frac{x^{4}}{4} + \frac{3 x^{2}}{2}$$

   Method #2

   1. Rewrite the integrand:

      $$x \left(x^{2} + 3\right) = x^{3} + 3 x$$

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

         $$\int 3 x\, dx = 3 \int x\, dx$$

         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: $\frac{3 x^{2}}{2}$

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

2. Now simplify:

   $$\frac{x^{2} \left(x^{2} + 6\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x^{2} \left(x^{2} + 6\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \left(x^{2} + 6\right)}{4}+ \mathrm{constant}$$