Step 1: Integral Simplification The integral is: \[ \int \frac{x^2 + 2x}{\sqrt{x^2 + 1}} \, dx \] Decompose into: \[ = \int \frac{x^2}{\sqrt{x^2 + 1}} \, dx + \int \frac{2x}{\sqrt{x^2 + 1}} \, dx \]
Step 2: Individual Integral Solution For the second integral, use substitution: \[ u = x^2 + 1 \quad \Rightarrow \quad du = 2x \, dx \] The integral becomes: \[ \int \frac{2x}{\sqrt{x^2 + 1}} \, dx = \int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{x^2 + 1} \] The first integral is simplified via substitution: \[ \int \frac{x^2}{\sqrt{x^2 + 1}} \, dx = \int \sqrt{x^2 + 1} \, dx \] This is solvable using standard methods for square root integration.
Answer: The solution is: \[ \frac{1}{2} \left( x^2 + 1 \right)^{3/2} + C \]
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]