Integral of sqrt(1+9/4x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \frac{9 x}{4} + 1$.

      Then let $du = \frac{9 dx}{4}$ and substitute $\frac{4 du}{9}$:

      $$\int \frac{4 \sqrt{u}}{9}\, du$$

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

         $$\int \sqrt{u}\, du = \frac{4 \int \sqrt{u}\, du}{9}$$

         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: $\frac{8 u^{\frac{3}{2}}}{27}$

      Now substitute $u$ back in:

      $$\frac{8 \left(\frac{9 x}{4} + 1\right)^{\frac{3}{2}}}{27}$$

   Method #2

   1. Rewrite the integrand:

      $$\text{True}$$

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

      $$\int \frac{\sqrt{9 x + 4}}{2}\, dx = \frac{\int \sqrt{9 x + 4}\, dx}{2}$$

      1. Let $u = 9 x + 4$.

         Then let $du = 9 dx$ and substitute $\frac{du}{9}$:

         $$\int \frac{\sqrt{u}}{9}\, du$$

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

            $$\int \sqrt{u}\, du = \frac{\int \sqrt{u}\, du}{9}$$

            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: $\frac{2 u^{\frac{3}{2}}}{27}$

         Now substitute $u$ back in:

         $$\frac{2 \left(9 x + 4\right)^{\frac{3}{2}}}{27}$$

      So, the result is: $\frac{\left(9 x + 4\right)^{\frac{3}{2}}}{27}$

2. Now simplify:

   $$\frac{\left(9 x + 4\right)^{\frac{3}{2}}}{27}$$

3. Add the constant of integration:

   $$\frac{\left(9 x + 4\right)^{\frac{3}{2}}}{27}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(9 x + 4\right)^{\frac{3}{2}}}{27}+ \mathrm{constant}$$