Final answer:
By using the comparison theorem and comparing the given function with 1/x², we can conclude that the integral ∫2∞(x - ln(x))/x¹dx converges since 1/x² converges and (x - ln(x))/x will eventually be less than 1/x².
Step-by-step explanation:
To determine whether the integral ∫2∞(x - ln(x))/x1dx converges or diverges, we can use the comparison theorem. The comparison theorem states that if 0 ≤ f(x) ≤ g(x) for all x in the interval [a, ∞), and if ∫a∞g(x)dx converges, then ∫a∞f(x)dx also converges. Conversely, if ∫a∞f(x)dx diverges, then ∫a∞g(x)dx also diverges. We will compare our function with 1/x2 as x approaches infinity.
Let's consider the function 1/x2. We know that ∫2∞1/x2dx is a p-integral with p > 1, which means it converges. Since ln(x) grows slower than x as x approaches infinity, the function (x - ln(x))/x will eventually be less than 1/x2 for sufficiently large x.
Therefore, by the comparison theorem, since (x - ln(x))/x is eventually less than a function that converges when integrated from 2 to infinity, we can conclude that ∫2∞(x - ln(x))/x1dx also converges.