1. **Problem Statement:** We want to find the integral $$\int \sqrt{1+u^2} \, du$$. 2. **Formula and Method:** This integral is a standard form that can be solved using integration by parts or recognizing it as a hyperbolic substitution integral. The formula for this integral is: $$\int \sqrt{1+u^2} \, du = \frac{u}{2} \sqrt{1+u^2} + \frac{1}{2} \ln \left| u + \sqrt{1+u^2} \right| + C$$ where $C$ is the constant of integration. 3. **Step-by-step Explanation:** - Let’s set $I = \int \sqrt{1+u^2} \, du$. - Use integration by parts: choose - $v = u$ so that $dv = du$ - $dw = \sqrt{1+u^2} \, du$ - Alternatively, use substitution $u = \sinh t$, then $du = \cosh t \, dt$ and $\sqrt{1+u^2} = \sqrt{1+\sinh^2 t} = \cosh t$. - The integral becomes: $$I = \int \cosh^2 t \, dt$$ - Use the identity $\cosh^2 t = \frac{1 + \cosh 2t}{2}$: $$I = \int \frac{1 + \cosh 2t}{2} \, dt = \frac{t}{2} + \frac{\sinh 2t}{4} + C$$ - Substitute back $t = \sinh^{-1} u$ and use $\sinh 2t = 2 \sinh t \cosh t = 2u \sqrt{1+u^2}$: $$I = \frac{\sinh^{-1} u}{2} + \frac{u \sqrt{1+u^2}}{2} + C$$ - Using the logarithmic form of inverse hyperbolic sine: $$\sinh^{-1} u = \ln \left| u + \sqrt{1+u^2} \right|$$ - So the final answer is: $$\int \sqrt{1+u^2} \, du = \frac{u}{2} \sqrt{1+u^2} + \frac{1}{2} \ln \left| u + \sqrt{1+u^2} \right| + C$$ 4. **Interpretation:** This integral represents the area under the curve of the function $f(u) = \sqrt{1+u^2}$, which grows faster than a linear function due to the $u^2$ term inside the square root. 5. **Graph Explanation:** The blue curve $f(x)$ described matches the shape of $\sqrt{1+x^2}$, starting near zero and increasing smoothly. The green curve $g(x)$ is different and not directly related to this integral. **Final answer:** $$\int \sqrt{1+u^2} \, du = \frac{u}{2} \sqrt{1+u^2} + \frac{1}{2} \ln \left| u + \sqrt{1+u^2} \right| + C$$