1. The problem is to evaluate the integral $$\int \sqrt{1+u^2} \, du$$. 2. To solve this, we use a trigonometric substitution. Let $$u = \sinh(t)$$, so that $$du = \cosh(t) \, dt$$. 3. Substitute into the integral: $$\int \sqrt{1+\sinh^2(t)} \cosh(t) \, dt$$. 4. Recall the identity $$1 + \sinh^2(t) = \cosh^2(t)$$, so the integral becomes: $$\int \sqrt{\cosh^2(t)} \cosh(t) \, dt = \int \cosh(t) \cosh(t) \, dt = \int \cosh^2(t) \, dt$$. 5. Use the identity $$\cosh^2(t) = \frac{1 + \cosh(2t)}{2}$$ to rewrite the integral: $$\int \frac{1 + \cosh(2t)}{2} \, dt = \frac{1}{2} \int 1 \, dt + \frac{1}{2} \int \cosh(2t) \, dt$$. 6. Integrate each term: $$\frac{1}{2} t + \frac{1}{2} \cdot \frac{\sinh(2t)}{2} + C = \frac{t}{2} + \frac{\sinh(2t)}{4} + C$$. 7. Substitute back to $$u$$: Since $$u = \sinh(t)$$, then $$t = \sinh^{-1}(u)$$. Also, $$\sinh(2t) = 2 \sinh(t) \cosh(t) = 2u \sqrt{1+u^2}$$. 8. Therefore, the integral is: $$\frac{1}{2} \sinh^{-1}(u) + \frac{1}{4} \cdot 2u \sqrt{1+u^2} + C = \frac{1}{2} \sinh^{-1}(u) + \frac{u}{2} \sqrt{1+u^2} + C$$. Final answer: $$\int \sqrt{1+u^2} \, du = \frac{u}{2} \sqrt{1+u^2} + \frac{1}{2} \sinh^{-1}(u) + C$$