1. **Problem:** Evaluate the integral $$\int \sqrt{x^2 + \pi} \, dx$$. 2. **Formula and rules:** This integral involves a square root of a quadratic expression. A common approach is to use a hyperbolic substitution or integration by parts. 3. **Step 1: Use substitution** Let $$x = \sqrt{\pi} \sinh t$$, then $$dx = \sqrt{\pi} \cosh t \, dt$$. 4. **Step 2: Simplify the integrand** $$\sqrt{x^2 + \pi} = \sqrt{\pi \sinh^2 t + \pi} = \sqrt{\pi (\sinh^2 t + 1)} = \sqrt{\pi} \cosh t$$. 5. **Step 3: Substitute into the integral** $$\int \sqrt{x^2 + \pi} \, dx = \int \sqrt{\pi} \cosh t \cdot \sqrt{\pi} \cosh t \, dt = \pi \int \cosh^2 t \, dt$$. 6. **Step 4: Use identity for $$\cosh^2 t$$** Recall $$\cosh^2 t = \frac{1 + \cosh 2t}{2}$$. 7. **Step 5: Integrate** $$\pi \int \cosh^2 t \, dt = \pi \int \frac{1 + \cosh 2t}{2} \, dt = \frac{\pi}{2} \int (1 + \cosh 2t) \, dt = \frac{\pi}{2} \left(t + \frac{\sinh 2t}{2}\right) + C$$. 8. **Step 6: Back-substitute $$t$$** Recall $$t = \sinh^{-1} \frac{x}{\sqrt{\pi}}$$ and $$\sinh 2t = 2 \sinh t \cosh t = 2 \frac{x}{\sqrt{\pi}} \frac{\sqrt{x^2 + \pi}}{\sqrt{\pi}} = \frac{2x \sqrt{x^2 + \pi}}{\pi}$$. 9. **Step 7: Final answer** $$\int \sqrt{x^2 + \pi} \, dx = \frac{\pi}{2} \sinh^{-1} \left(\frac{x}{\sqrt{\pi}}\right) + \frac{x \sqrt{x^2 + \pi}}{2} + C$$. This completes the solution.