Integral of sqrt^8(x-1)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sqrt{x - 1}$.

      Then let $du = \frac{dx}{2 \sqrt{x - 1}}$ and substitute $2 du$:

      $$\int 2 u^{9}\, du$$

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

         $$\int u^{9}\, du = 2 \int u^{9}\, du$$

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

            $$\int u^{9}\, du = \frac{u^{10}}{10}$$

         So, the result is: $\frac{u^{10}}{5}$

      Now substitute $u$ back in:

      $$\frac{\left(x - 1\right)^{5}}{5}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(\sqrt{x - 1}\right)^{8} = x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1$$

   2. Integrate term-by-term:

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

         $$\int x^{4}\, dx = \frac{x^{5}}{5}$$

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

         $$\int \left(- 4 x^{3}\right)\, dx = - 4 \int x^{3}\, dx$$

         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: $- x^{4}$

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

         $$\int 6 x^{2}\, dx = 6 \int x^{2}\, dx$$

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

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

         So, the result is: $2 x^{3}$

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

         $$\int \left(- 4 x\right)\, dx = - 4 \int x\, dx$$

         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: $- 2 x^{2}$

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

         $$\int 1\, dx = x$$

      The result is: $\frac{x^{5}}{5} - x^{4} + 2 x^{3} - 2 x^{2} + x$

2. Now simplify:

   $$\frac{\left(x - 1\right)^{5}}{5}$$

3. Add the constant of integration:

   $$\frac{\left(x - 1\right)^{5}}{5}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(x - 1\right)^{5}}{5}+ \mathrm{constant}$$