Integral of sqrt(1+8x) dx

1. Let $u = 8 x + 1$.

   Then let $du = 8 dx$ and substitute $\frac{du}{8}$:

   $$\int \frac{\sqrt{u}}{8}\, du$$

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

      $$\int \sqrt{u}\, du = \frac{\int \sqrt{u}\, du}{8}$$

      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: $\frac{u^{\frac{3}{2}}}{12}$

   Now substitute $u$ back in:

   $$\frac{\left(8 x + 1\right)^{\frac{3}{2}}}{12}$$

2. Add the constant of integration:

   $$\frac{\left(8 x + 1\right)^{\frac{3}{2}}}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(8 x + 1\right)^{\frac{3}{2}}}{12}+ \mathrm{constant}$$