Integral of 1/x^1/2*(x-1) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

      1. Integrate term-by-term:

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

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

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

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

            $$\int \left(-2\right)\, du = - 2 u$$

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

      Now substitute $u$ back in:

      $$\frac{2 x^{\frac{3}{2}}}{3} - 2 \sqrt{x}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{x - 1}{\sqrt{x}} = \frac{x}{\sqrt{x}} - \frac{1}{\sqrt{x}}$$

   2. Integrate term-by-term:

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

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

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

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

            $$\int \frac{1}{u^{4}}\, du = - 2 \int \frac{1}{u^{4}}\, du$$

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

               $$\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$$

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

         Now substitute $u$ back in:

         $$\frac{2 x^{\frac{3}{2}}}{3}$$

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

         $$\int \left(- \frac{1}{\sqrt{x}}\right)\, dx = - \int \frac{1}{\sqrt{x}}\, dx$$

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

            $$\int \frac{1}{\sqrt{x}}\, dx = 2 \sqrt{x}$$

         So, the result is: $- 2 \sqrt{x}$

      The result is: $\frac{2 x^{\frac{3}{2}}}{3} - 2 \sqrt{x}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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