1. **State the problem:** We need to evaluate the integral $$\int \sqrt{1 + y^2} \, dy$$. 2. **Formula and approach:** This is an integral involving a square root of a quadratic expression. A common method is to use a hyperbolic substitution or recognize it as a standard integral form. 3. **Substitution:** Let’s use the substitution $$y = \sinh(t)$$, so that $$dy = \cosh(t) \, dt$$. 4. **Rewrite the integrand:** Note that $$1 + y^2 = 1 + \sinh^2(t) = \cosh^2(t)$$, so $$\sqrt{1 + y^2} = \cosh(t)$$. 5. **Substitute into the integral:** $$\int \sqrt{1 + y^2} \, dy = \int \cosh(t) \cdot \cosh(t) \, dt = \int \cosh^2(t) \, dt$$. 6. **Use identity:** Recall $$\cosh^2(t) = \frac{1 + \cosh(2t)}{2}$$. 7. **Integrate:** $$\int \cosh^2(t) \, dt = \int \frac{1 + \cosh(2t)}{2} \, dt = \frac{1}{2} \int 1 \, dt + \frac{1}{2} \int \cosh(2t) \, dt = \frac{t}{2} + \frac{1}{4} \sinh(2t) + C$$. 8. **Back-substitute:** Since $$y = \sinh(t)$$, then $$t = \sinh^{-1}(y)$$ and $$\sinh(2t) = 2 \sinh(t) \cosh(t) = 2 y \sqrt{1 + y^2}$$. 9. **Final answer:** $$\int \sqrt{1 + y^2} \, dy = \frac{1}{2} \sinh^{-1}(y) + \frac{y}{2} \sqrt{1 + y^2} + C$$. This is the antiderivative of the given integral.