Integral of (1-sqrt(y-2))^2 dy

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sqrt{y - 2}$.

      Then let $du = \frac{dy}{2 \sqrt{y - 2}}$ and substitute $du$:

      $$\int \left(2 u^{3} - 4 u^{2} + 2 u\right)\, du$$

      1. Integrate term-by-term:

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

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

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

               $$\int u^{3}\, du = \frac{u^{4}}{4}$$

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

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

            $$\int \left(- 4 u^{2}\right)\, du = - 4 \int u^{2}\, du$$

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

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

            So, the result is: $- \frac{4 u^{3}}{3}$

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

            $$\int 2 u\, du = 2 \int u\, du$$

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

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

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

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

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

         1. Let $u = y - 2$.

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

            $$\int \sqrt{u}\, du$$

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

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

            Now substitute $u$ back in:

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

         So, the result is: $- \frac{4 \left(y - 2\right)^{\frac{3}{2}}}{3}$

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

         $$\int \left(-1\right)\, dy = - y$$

      The result is: $\frac{y^{2}}{2} - y - \frac{4 \left(y - 2\right)^{\frac{3}{2}}}{3}$

   Method #3

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

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

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

         1. Let $u = y - 2$.

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

            $$\int \sqrt{u}\, du$$

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

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

            Now substitute $u$ back in:

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

         So, the result is: $- \frac{4 \left(y - 2\right)^{\frac{3}{2}}}{3}$

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

         $$\int \left(-1\right)\, dy = - y$$

      The result is: $\frac{y^{2}}{2} - y - \frac{4 \left(y - 2\right)^{\frac{3}{2}}}{3}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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