Integral of 6x(3x^2+1)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 3 x^{2} + 1$.

      Then let $du = 6 x dx$ and substitute $du$:

      $$\int u\, du$$

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

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

      Now substitute $u$ back in:

      $$\frac{\left(3 x^{2} + 1\right)^{2}}{2}$$

   Method #2

   1. Rewrite the integrand:

      $$6 x \left(3 x^{2} + 1\right) = 18 x^{3} + 6 x$$

   2. Integrate term-by-term:

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

         $$\int 18 x^{3}\, dx = 18 \int x^{3}\, dx$$

         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: $\frac{9 x^{4}}{2}$

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

         $$\int 6 x\, dx = 6 \int x\, dx$$

         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: $3 x^{2}$

      The result is: $\frac{9 x^{4}}{2} + 3 x^{2}$

2. Now simplify:

   $$\frac{\left(3 x^{2} + 1\right)^{2}}{2}$$

3. Add the constant of integration:

   $$\frac{\left(3 x^{2} + 1\right)^{2}}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(3 x^{2} + 1\right)^{2}}{2}+ \mathrm{constant}$$