If $f'(x) = \frac{1}{-x + \sqrt{x^2 + 1}}$ and $f(0) = -\frac{1 + \sqrt{2}}{2}$, then the value of $f(5)$ will be :
Step-by-Step Solution
Key Concept: Rationalize the denominator of f'(x) by multiplying by the conjugate (x + √(x² + 1))/(x + √(x² + 1)) to convert it into an integrable form, then use the standard integral ∫√(x² + 1)dx = (x√(x² + 1))/2 + (1/2)ln|x + √(x² + 1)| + C with the initial condition f(0) = -(1+√2)/2 to find the constant and evaluate f(5).
Computing $f(x) = \int\left(x + \sqrt{x^2 + 1}\right) dx = \frac{x^2}{2} + \frac{x\sqrt{x^2+1}}{2} + \frac{1}{2}\log\left|x + \sqrt{x^2+1}\right| + c$. Using the boundary condition $f(0) = c$ and the given value $f(0) = -\frac{1+\sqrt{2}}{2}$, the constant of integration is determined.
Correct Answer: 1,3