Integral of (6x)/(sqrt(4x+5)) dx

1. Let $u = \sqrt{4 x + 5}$.

   Then let $du = \frac{2 dx}{\sqrt{4 x + 5}}$ and substitute $du$:

   $$\int \left(\frac{3 u^{2}}{4} - \frac{15}{4}\right)\, du$$

   1. Integrate term-by-term:

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

         $$\int \frac{3 u^{2}}{4}\, du = \frac{3 \int u^{2}\, du}{4}$$

         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: $\frac{u^{3}}{4}$

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

         $$\int \left(- \frac{15}{4}\right)\, du = - \frac{15 u}{4}$$

      The result is: $\frac{u^{3}}{4} - \frac{15 u}{4}$

   Now substitute $u$ back in:

   $$\frac{\left(4 x + 5\right)^{\frac{3}{2}}}{4} - \frac{15 \sqrt{4 x + 5}}{4}$$

2. Now simplify:

   $$\left(x - \frac{5}{2}\right) \sqrt{4 x + 5}$$

3. Add the constant of integration:

   $$\left(x - \frac{5}{2}\right) \sqrt{4 x + 5}+ \mathrm{constant}$$

The answer is: