Integral of (y-1)^2 dy

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = y - 1$.

      Then let $du = dy$ and substitute $du$:

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

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

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

      Now substitute $u$ back in:

      $$\frac{\left(y - 1\right)^{3}}{3}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(y - 1\right)^{2} = y^{2} - 2 y + 1$$

   2. Integrate term-by-term:

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

         $$\int y^{2}\, dy = \frac{y^{3}}{3}$$

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

         $$\int \left(- 2 y\right)\, dy = - 2 \int y\, dy$$

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

            $$\int y\, dy = \frac{y^{2}}{2}$$

         So, the result is: $- y^{2}$

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

         $$\int 1\, dy = y$$

      The result is: $\frac{y^{3}}{3} - y^{2} + y$

2. Now simplify:

   $$\frac{\left(y - 1\right)^{3}}{3}$$

3. Add the constant of integration:

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

The answer is:

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