1. **Problem:** Determine whether the integral $\int_0^\infty \frac{1}{\sqrt{x} (1 + x)} \, dx$ is convergent or divergent and evaluate it if convergent. 2. **Formula and rules:** We analyze the integral by checking behavior near 0 and $\infty$. The integral is improper at both limits. 3. **Step 1: Behavior near 0** As $x \to 0^+$, $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{\sqrt{x}}$ since $1+x \approx 1$. The integral near 0 behaves like $\int_0 \frac{1}{\sqrt{x}} dx = \int_0 x^{-1/2} dx$, which converges because $\int_0^a x^p dx$ converges if $p > -1$. 4. **Step 2: Behavior near $\infty$** As $x \to \infty$, $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{\sqrt{x} \cdot x} = \frac{1}{x^{3/2}}$. The integral near infinity behaves like $\int^\infty \frac{1}{x^{3/2}} dx$, which converges because $\int^\infty x^{-p} dx$ converges if $p > 1$. 5. **Step 3: Evaluate the integral** Rewrite the integral: $$\int_0^\infty \frac{1}{\sqrt{x}(1+x)} dx = \int_0^\infty \frac{1}{x^{1/2}(1+x)} dx$$ Use substitution $x = t^2$, so $dx = 2t dt$, $\sqrt{x} = t$. Then: $$\int_0^\infty \frac{1}{t (1 + t^2)} 2t dt = 2 \int_0^\infty \frac{1}{1 + t^2} dt$$ Simplify: $$2 \int_0^\infty \frac{1}{1 + t^2} dt$$ 6. **Step 4: Evaluate the integral** We know: $$\int_0^\infty \frac{1}{1 + t^2} dt = \frac{\pi}{2}$$ Therefore: $$2 \times \frac{\pi}{2} = \pi$$ 7. **Conclusion:** The integral converges and its value is $\pi$. --- **Final answer:** $$\int_0^\infty \frac{1}{\sqrt{x} (1 + x)} dx = \pi$$