Integral of sqrtsqrtx/1+sqrtx dx

1. Integrate term-by-term:

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

      $$\int \frac{\sqrt{\sqrt{x}}}{1}\, dx = \int \sqrt{\sqrt{x}}\, dx$$

      1. Don't know the steps in finding this integral.

         But the integral is

         $$\frac{4 x^{\frac{5}{4}}}{5}$$

      So, the result is: $\frac{4 x^{\frac{5}{4}}}{5}$

   1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int \sqrt{x}\, dx = \frac{2 x^{\frac{3}{2}}}{3}$$

   The result is: $\frac{4 x^{\frac{5}{4}}}{5} + \frac{2 x^{\frac{3}{2}}}{3}$

2. Add the constant of integration:

   $$\frac{4 x^{\frac{5}{4}}}{5} + \frac{2 x^{\frac{3}{2}}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{4 x^{\frac{5}{4}}}{5} + \frac{2 x^{\frac{3}{2}}}{3}+ \mathrm{constant}$$