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

1. Let $u = \sqrt{2 x + 1}$.

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

   $$\int \left(\frac{u^{2}}{2} - \frac{1}{2}\right)\, du$$

   1. Integrate term-by-term:

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

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

         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: $\frac{u^{3}}{6}$

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

         $$\int \left(- \frac{1}{2}\right)\, du = - \frac{u}{2}$$

      The result is: $\frac{u^{3}}{6} - \frac{u}{2}$

   Now substitute $u$ back in:

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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