Integral of x/sqrt(4x+1) dx

1. Let $u = \sqrt{4 x + 1}$.

   Then let $du = \frac{2 dx}{\sqrt{4 x + 1}}$ and substitute $du$:

   $$\int \left(\frac{u^{2}}{8} - \frac{1}{8}\right)\, du$$

   1. Integrate term-by-term:

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

         $$\int \frac{u^{2}}{8}\, du = \frac{\int u^{2}\, du}{8}$$

         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: $\frac{u^{3}}{24}$

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

         $$\int \left(- \frac{1}{8}\right)\, du = - \frac{u}{8}$$

      The result is: $\frac{u^{3}}{24} - \frac{u}{8}$

   Now substitute $u$ back in:

   $$\frac{\left(4 x + 1\right)^{\frac{3}{2}}}{24} - \frac{\sqrt{4 x + 1}}{8}$$

2. Now simplify:

   $$\frac{\left(2 x - 1\right) \sqrt{4 x + 1}}{12}$$

3. Add the constant of integration:

   $$\frac{\left(2 x - 1\right) \sqrt{4 x + 1}}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(2 x - 1\right) \sqrt{4 x + 1}}{12}+ \mathrm{constant}$$