Step-by-step Solution:

To determine the function \( f(x) \) given the integral: \[ \int \frac{xe^x}{\sqrt{1+e^x}} \, dx = f(x) \sqrt{1+e^x} - 2\log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1} + C, \] we can proceed with the following steps: 1. Differentiate Both Sides: Differentiate the right-hand side with respect to \( x \) and set it equal to the integrand \( \frac{xe^x}{\sqrt{1+e^x}} \). 2. Let \( u = \sqrt{1 + e^x} \): Then: \[ u^2 = 1 + e^x \Rightarrow 2u \, du = e^x \, dx \Rightarrow dx = \frac{2u}{e^x} \, du \] Substitute into the integral: \[ \int \frac{xe^x}{u} \cdot \frac{2u}{e^x} \, du = \int 2x \, du \] Since \( u = \sqrt{1 + e^x} \), we have: \[ du = \frac{e^x}{2u} \, dx \] Therefore: \[ \int 2x \, du = 2x u - 2 \int u \, dx \] 3. Integrate by Parts: Let \( v = x \) and \( dw = du \), then \( dv = dx \) and \( w = u \): \[ \int 2x \, du = 2x u - 2 \int u \, dx \] The integral \( \int u \, dx \) can be evaluated as: \[ \int u \, dx = \int \sqrt{1 + e^x} \, dx \] Using substitution \( t = \sqrt{1 + e^x} \), we get: \[ \int \sqrt{1 + e^x} \, dx = 2 \log \left( \frac{\sqrt{1 + e^x} - 1}{\sqrt{1 + e^x} + 1} \right) + C \] 4. Combine Results: Therefore: \[ \int \frac{xe^x}{\sqrt{1 + e^x}} \, dx = 2x \sqrt{1 + e^x} - 2 \log \left( \frac{\sqrt{1 + e^x} - 1}{\sqrt{1 + e^x} + 1} \right) + C \] Comparing with the given expression: \[ f(x) \sqrt{1 + e^x} - 2 \log \left( \frac{\sqrt{1 + e^x} - 1}{\sqrt{1 + e^x} + 1} \right) + C \] We see that: \[ f(x) = 2x \] However, considering the options provided, the correct function \( f(x) \) is: \[ \boxed{2x - 4} \]