Find the continuous function $ f(x)$ such that $ xf(x)\minus{}\int_0^x f(t)\ dt\equal{}x\plus{}\ln (\sqrt{x^2\plus{}1}\minus{}x)$ with $ f(0)\equal{}\ln 2$.
1. Given the equation:
   \[
   x f(x) - \int_0^x f(t) \, dt = x + \ln (\sqrt{x^2 + 1} - x)
   \]
   with the initial condition \( f(0) = \ln 2 \).

2. Differentiate both sides with respect to \( x \):
   \[
   \frac{d}{dx} \left( x f(x) - \int_0^x f(t) \, dt \right) = \frac{d}{dx} \left( x + \ln (\sqrt{x^2 + 1} - x) \right)
   \]

3. Using the Leibniz rule for differentiation under the integral sign, we get:
   \[
   \frac{d}{dx} \left( x f(x) \right) - \frac{d}{dx} \left( \int_0^x f(t) \, dt \right) = 1 + \frac{d}{dx} \left( \ln (\sqrt{x^2 + 1} - x) \right)
   \]
   \[
   f(x) + x f'(x) - f(x) = 1 + \frac{d}{dx} \left( \ln (\sqrt{x^2 + 1} - x) \right)
   \]

4. Simplify the left-hand side:
   \[
   x f'(x) = 1 + \frac{d}{dx} \left( \ln (\sqrt{x^2 + 1} - x) \right)
   \]

5. Compute the derivative on the right-hand side:
   \[
   \frac{d}{dx} \left( \ln (\sqrt{x^2 + 1} - x) \right) = \frac{1}{\sqrt{x^2 + 1} - x} \cdot \frac{d}{dx} (\sqrt{x^2 + 1} - x)
   \]
   \[
   = \frac{1}{\sqrt{x^2 + 1} - x} \left( \frac{x}{\sqrt{x^2 + 1}} - 1 \right)
   \]
   \[
   = \frac{x - \sqrt{x^2 + 1}}{(\sqrt{x^2 + 1} - x) \sqrt{x^2 + 1}}
   \]
   \[
   = -\frac{1}{\sqrt{x^2 + 1}}
   \]

6. Substitute back into the equation:
   \[
   x f'(x) = 1 - \frac{1}{\sqrt{x^2 + 1}}
   \]
   \[
   f'(x) = \frac{1 - \frac{1}{\sqrt{x^2 + 1}}}{x}
   \]

7. Integrate both sides with respect to \( x \):
   \[
   f(x) = \int \frac{1 - \frac{1}{\sqrt{x^2 + 1}}}{x} \, dx
   \]

8. Simplify the integrand:
   \[
   f(x) = \int \left( \frac{1}{x} - \frac{1}{x \sqrt{x^2 + 1}} \right) \, dx
   \]

9. Integrate each term separately:
   \[
   \int \frac{1}{x} \, dx = \ln |x|
   \]
   \[
   \int \frac{1}{x \sqrt{x^2 + 1}} \, dx = \ln \left( 1 + \sqrt{x^2 + 1} \right)
   \]

10. Combine the results:
    \[
    f(x) = \ln |x| - \ln \left( 1 + \sqrt{x^2 + 1} \right) + C
    \]

11. Use the initial condition \( f(0) = \ln 2 \):
    \[
    \ln 2 = \ln (1 + \sqrt{1}) + C = \ln 2 + C
    \]
    \[
    C = 0
    \]

12. Therefore, the function is:
    \[
    f(x) = \ln (1 + \sqrt{x^2 + 1})
    \]

The final answer is \( \boxed{ f(x) = \ln (1 + \sqrt{x^2 + 1}) } \).