1. **State the problem:** Evaluate the integral $$\int \frac{\sqrt{x}}{x^2+1} \, dx.$$\n\n2. **Rewrite the integrand:** Note that $$\sqrt{x} = x^{\frac{1}{2}}.$$ So the integral becomes $$\int \frac{x^{\frac{1}{2}}}{x^2+1} \, dx.$$\n\n3. **Consider substitution:** Let $$t = x^{\frac{1}{2}} = \sqrt{x}.$$ Then $$x = t^2$$ and $$dx = 2t \, dt.$$\n\n4. **Rewrite the integral in terms of $$t$$:** Substitute into the integral:\n$$\int \frac{t}{(t^2)^2 + 1} \cdot 2t \, dt = \int \frac{2t^2}{t^4 + 1} \, dt.$$\n\n5. **Simplify the integral:**\n$$\int \frac{2t^2}{t^4 + 1} \, dt.$$\n\n6. **Use symmetry in the denominator:** The denominator $$t^4 + 1$$ can be factored as $$(t^2 - \sqrt{2}t + 1)(t^2 + \sqrt{2}t + 1)$$ but partial fractions are complicated here. Instead, use the substitution $$t = \tan \theta$$ or try a different approach.\n\n7. **Alternative substitution:** Let $$u = t^2,$$ then $$du = 2t \, dt,$$ which matches the numerator differential. So rewrite integral as:\n$$\int \frac{2t^2}{t^4 + 1} \, dt = \int \frac{u}{u^2 + 1} \, du.$$\n\n8. **Integral becomes:**\n$$\int \frac{u}{u^2 + 1} \, du.$$\n\n9. **Use substitution for this integral:** Let $$w = u^2 + 1,$$ then $$dw = 2u \, du,$$ so $$u \, du = \frac{dw}{2}.$$\n\n10. **Rewrite integral:**\n$$\int \frac{u}{u^2 + 1} \, du = \int \frac{1}{w} \cdot u \, du = \int \frac{1}{w} \cdot \frac{dw}{2} = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln|u^2 + 1| + C.$$\n\n11. **Back-substitute:** Recall $$u = t^2$$ and $$t = \sqrt{x}$$, so\n$$u^2 + 1 = (t^2)^2 + 1 = x^2 + 1.$$\n\n12. **Final answer:**\n$$\int \frac{\sqrt{x}}{x^2 + 1} \, dx = \frac{1}{2} \ln(x^2 + 1) + C.$$