To solve the given integral \(\int_{0}^{1}\int_{x}^{1} \frac{1}{1+y^2} \mathrm{dy}\mathrm{dx}\), we will use the method of iterated integrals focusing on each variable independently:

1. Consider the inner integral first, which involves the variable \(y\): \(\int_{x}^{1} \frac{1}{1+y^2} \mathrm{dy}\)
2. The integral of \(\frac{1}{1+y^2}\) is known to be \(\tan^{-1}(y)\). Therefore, the integral becomes: \(\left[ \tan^{-1}(y) \right]_{x}^{1}\)
3. Evaluate this definite integral:
   • Substitute the upper limit \((y=1)\): \(\tan^{-1}(1) = \frac{\pi}{4}\)
   • Substitute the lower limit \((y=x)\): \(\tan^{-1}(x)\)
4. Substitute this result into the outer integral: \(\int_{0}^{1} \left(\frac{\pi}{4} - \tan^{-1}(x)\right) \mathrm{dx}\)
5. Break this into two separate integrals: \(\int_{0}^{1} \frac{\pi}{4} \, \mathrm{dx} - \int_{0}^{1} \tan^{-1}(x) \, \mathrm{dx}\)
6. Evaluate the first integral:
   • \(\int_{0}^{1} \frac{\pi}{4} \, \mathrm{dx} = \frac{\pi}{4} \times [x]_{0}^{1} = \frac{\pi}{4}\times(1-0) = \frac{\pi}{4}\)
7. For the second integral, use the integration by parts method:
   • Let \(u = \tan^{-1}(x)\); \(\mathrm{dv} = \mathrm{dx}\)
   • Then, \(\mathrm{du} = \frac{1}{1+x^2} \, \mathrm{dx}\); \(v = x\)
   • Here: \([\tan^{-1}(x) \cdot x]_{0}^{1} - \int_{0}^{1} x \cdot \frac{1}{1+x^2} \, \mathrm{dx}\)
   • Evaluate: \(\left[\tan^{-1}(1) \cdot 1 - \tan^{-1}(0) \cdot 0 - \int_{0}^{1} \frac{x}{1+x^2} \, \mathrm{dx}\right]\)
   • Solve remaining integral \(\int_{0}^{1} \frac{x}{1+x^2} \, \mathrm{dx}\) by setting \(w = 1+x^2 \Rightarrow \mathrm{dw} = 2x \, \mathrm{dx}\), here \(\frac{1}{2}\int \frac{1}{w} \, \mathrm{dw} = \frac{1}{2} \ln|w| \Bigg|_0^1 = \frac{1}{2} \ln(2)\)
8. The final expression simplifies: \(\frac{\pi}{4} - \left( \frac{\pi}{4} - \frac{1}{2} \ln(2) \right) = \frac{1}{2} \ln(2)\)
9. Therefore, the value of the integral is \(\frac{1}{2} \ln(2)\).

Therefore, the correct answer is: \(\frac{1}{2} \ln(2)\).