Integral of 1/1+(sqrt(2x-1)) dx

1. Integrate term-by-term:

   1. Let $u = 2 x - 1$.

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

      $$\int \frac{\sqrt{u}}{2}\, du$$

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

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

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

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

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

      Now substitute $u$ back in:

      $$\frac{\left(2 x - 1\right)^{\frac{3}{2}}}{3}$$

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

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

   The result is: $x + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{3}$

2. Now simplify:

   $$x + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{3}$$

3. Add the constant of integration:

   $$x + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$x + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{3}+ \mathrm{constant}$$