1. **Stating the problem:** Evaluate the integral $$\int \frac{\sqrt{\sqrt{x}}}{x^2+1} \, dx$$. 2. **Rewrite the integrand:** Note that $$\sqrt{\sqrt{x}} = (x^{1/2})^{1/2} = x^{1/4}$$. So the integral becomes $$\int \frac{x^{1/4}}{x^2+1} \, dx$$. 3. **Substitution attempt:** Let us try substitution to simplify the integral. Set $$t = x^{1/4}$$, so $$x = t^4$$. 4. **Find differential:** Then $$dx = 4t^3 \, dt$$. 5. **Rewrite the integral in terms of t:** $$\int \frac{t}{(t^4)^2 + 1} \cdot 4t^3 \, dt = \int \frac{t}{t^8 + 1} \cdot 4t^3 \, dt = \int \frac{4t^4}{t^8 + 1} \, dt$$. 6. **Simplify the integral:** $$\int \frac{4t^4}{t^8 + 1} \, dt$$. 7. **Recognize the integral form:** The denominator is $$t^8 + 1$$, which can be factored but is complicated. This integral does not simplify easily with elementary functions. 8. **Conclusion:** The integral $$\int \frac{\sqrt{\sqrt{x}}}{x^2+1} \, dx$$ can be rewritten as $$\int \frac{4t^4}{t^8 + 1} \, dt$$ with $$t = x^{1/4}$$, but it does not have a simple elementary antiderivative. **Final answer:** $$\int \frac{\sqrt{\sqrt{x}}}{x^2+1} \, dx = \int \frac{4t^4}{t^8 + 1} \, dt \quad \text{where} \quad t = x^{1/4} + C$$