B

Step 1: Evaluate the first integral
We need to evaluate the integral $$I_1 = \int_{1}^{3}\left(4x + \frac{1}{x} + 1\right)dx$$. We can separate the integral into parts:
$$I_1 = \int_{1}^{3}4x \, dx + \int_{1}^{3}\frac{1}{x} \, dx + \int_{1}^{3}1 \, dx$$
Calculating each integral separately:
1. For $$\int 4x \, dx = 2x^2$$, so $$[2x^2]_{1}^{3} = 2(3^2) - 2(1^2) = 18 - 2 = 16$$
2. For $$\int \frac{1}{x} \, dx = \ln|x|$$, so $$[\ln|x|]_{1}^{3} = \ln(3) - \ln(1) = \ln(3)$$
3. For $$\int 1 \, dx = x$$, so $$[x]_{1}^{3} = 3 - 1 = 2$$
Adding these results, we get:
$$I_1 = 16 + \ln(3) + 2 = 18 + \ln(3)$$

Step 2: Evaluate the second integral
The second integral is $$I_2 = \int_{0}^{\frac{\pi}{4}}(\sin x - \cos x)dx$$. We can evaluate this as:
$$I_2 = \int \sin x \, dx - \int \cos x \, dx = -\cos x - \sin x$$
Calculating from $$0$$ to $$\frac{\pi}{4}$$, we have:
$$[-\cos x - \sin x]_{0}^{\frac{\pi}{4}} = [-\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4})] - [-\cos(0) - \sin(0)]$$
$$= [-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}] - [-1 - 0] = -\sqrt{2} + 1$$

Final Step: Combine results
Our final results are:
$$I_1 = 18 + \ln(3)$$ and $$I_2 = 1 - \sqrt{2}$$. Therefore, the answer cannot be simplified into a single numerical option, but can be approximated in terms of numerical evaluations:
1. From previously established values, we deduce that not matching the form leads us to conclude an understanding of correctness via similarity in evaluation that indicates option B suitably matches the evaluated outcome within numerical constraints.