Integral of (1-x^2)3x^3dx 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 $du$:

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

      1. Integrate term-by-term:

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

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

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

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

            So, the result is: $\frac{u^{3}}{2}$

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

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

            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: $\frac{3 u^{2}}{4}$

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

      Now substitute $u$ back in:

      $$- \frac{x^{6}}{2} + \frac{3 x^{4}}{4}$$

   Method #2

   1. Rewrite the integrand:

      $$x^{3} \cdot 3 \left(1 - x^{2}\right) = - 3 x^{5} + 3 x^{3}$$

   2. Integrate term-by-term:

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

         $$\int \left(- 3 x^{5}\right)\, dx = - 3 \int x^{5}\, dx$$

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

            $$\int x^{5}\, dx = \frac{x^{6}}{6}$$

         So, the result is: $- \frac{x^{6}}{2}$

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

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

         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: $\frac{3 x^{4}}{4}$

      The result is: $- \frac{x^{6}}{2} + \frac{3 x^{4}}{4}$

2. Now simplify:

   $$\frac{x^{4} \left(3 - 2 x^{2}\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x^{4} \left(3 - 2 x^{2}\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{4} \left(3 - 2 x^{2}\right)}{4}+ \mathrm{constant}$$