Evaluate the integral: \[ \int \frac{dx}{(x+1)\sqrt{x^2 + 4}} \]
\( \frac{1}{\sqrt{5}} \cosh^{-1} \left( \frac{4+x}{2(x-1)} \right) + C \)
Step 1: Using a Suitable Substitution Let: \[ x = 2 \sinh t \] Differentiating both sides: \[ dx = 2\cosh t \, dt \] Rewriting the denominator: \[ \sqrt{x^2 + 4} = \sqrt{4\sinh^2 t + 4} = \sqrt{4(\sinh^2 t + 1)} = \sqrt{4\cosh^2 t} = 2\cosh t \] Thus, the integral transforms into: \[ \int \frac{dx}{(x+1)\sqrt{x^2+4}} = \int \frac{2\cosh t \, dt}{(2\sinh t + 1)2\cosh t} \]
Step 2: Evaluating the Integral The fraction simplifies to: \[ \int \frac{dt}{2\sinh t + 1} \] Using the inverse hyperbolic sine transformation: \[ I = \frac{1}{\sqrt{5}} \sinh^{-1} \left( \frac{4-x}{2(x+1)} \right) + C \]
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]

