1. The problem asks us to evaluate the expression $$\int_1^{\pi} 4x \, dx - \int_1^{\pi} \tan x \, dx.$$ 2. First, calculate $$\int_1^{\pi} 4x \, dx.$$ The antiderivative of $$4x$$ is $$2x^2$$ since $$\frac{d}{dx}(2x^2) = 4x$$. 3. Evaluate this definite integral: $$\int_1^{\pi} 4x \, dx = [2x^2]_1^{\pi} = 2\pi^2 - 2(1)^2 = 2\pi^2 - 2.$$ 4. Next, calculate $$\int_1^{\pi} \tan x \, dx.$$ The antiderivative of $$\tan x$$ is $$-\ln|\cos x|$$ because $$\frac{d}{dx}(-\ln|\cos x|) = \tan x$$. 5. Evaluate this definite integral: $$\int_1^{\pi} \tan x \, dx = [-\ln|\cos x|]_1^{\pi} = -\ln|\cos \pi| + \ln|\cos 1| = -\ln| -1 | + \ln|\cos 1| = 0 + \ln|\cos 1| = \ln(\cos 1).$$ Note that $$|\cos \pi| = |-1|=1$$ and $$\ln 1 =0$$. 6. Combine the results: $$\int_1^{\pi} 4x \, dx - \int_1^{\pi} \tan x \, dx = (2\pi^2 - 2) - \ln(\cos 1).$$ 7. Final simplified answer: $$\boxed{2\pi^2 - 2 - \ln(\cos 1)}.$$