Integral of x^2(1-x^3)^5 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 1 - x^{3}$.

      Then let $du = - 3 x^{2} dx$ and substitute $- \frac{du}{3}$:

      $$\int \left(- \frac{u^{5}}{3}\right)\, du$$

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

         $$\int u^{5}\, du = - \frac{\int u^{5}\, du}{3}$$

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

            $$\int u^{5}\, du = \frac{u^{6}}{6}$$

         So, the result is: $- \frac{u^{6}}{18}$

      Now substitute $u$ back in:

      $$- \frac{\left(1 - x^{3}\right)^{6}}{18}$$

   Method #2

   1. Rewrite the integrand:

      $$x^{2} \left(1 - x^{3}\right)^{5} = - x^{17} + 5 x^{14} - 10 x^{11} + 10 x^{8} - 5 x^{5} + x^{2}$$

   2. Integrate term-by-term:

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

         $$\int \left(- x^{17}\right)\, dx = - \int x^{17}\, dx$$

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

            $$\int x^{17}\, dx = \frac{x^{18}}{18}$$

         So, the result is: $- \frac{x^{18}}{18}$

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

         $$\int 5 x^{14}\, dx = 5 \int x^{14}\, dx$$

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

            $$\int x^{14}\, dx = \frac{x^{15}}{15}$$

         So, the result is: $\frac{x^{15}}{3}$

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

         $$\int \left(- 10 x^{11}\right)\, dx = - 10 \int x^{11}\, dx$$

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

            $$\int x^{11}\, dx = \frac{x^{12}}{12}$$

         So, the result is: $- \frac{5 x^{12}}{6}$

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

         $$\int 10 x^{8}\, dx = 10 \int x^{8}\, dx$$

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

            $$\int x^{8}\, dx = \frac{x^{9}}{9}$$

         So, the result is: $\frac{10 x^{9}}{9}$

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

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

         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: $- \frac{5 x^{6}}{6}$

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

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

      The result is: $- \frac{x^{18}}{18} + \frac{x^{15}}{3} - \frac{5 x^{12}}{6} + \frac{10 x^{9}}{9} - \frac{5 x^{6}}{6} + \frac{x^{3}}{3}$

2. Now simplify:

   $$- \frac{\left(x^{3} - 1\right)^{6}}{18}$$

3. Add the constant of integration:

   $$- \frac{\left(x^{3} - 1\right)^{6}}{18}+ \mathrm{constant}$$

The answer is:

$$- \frac{\left(x^{3} - 1\right)^{6}}{18}+ \mathrm{constant}$$