Integral of sqrt(1+9x) dx

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

   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:

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

2. Add the constant of integration:

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

The answer is:

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