Question

Use the Comparison Theorem of Section 7.8 to determine whether each of the following integrals converges or diverges.

(a) ∫(x/(x³ + 1)) dx
(b) ∫(1/(x ln x)) dx

Answer 1

The integral of x/(x³ + 1) converges, while the integral of 1/(x ln x) diverges according to the Comparison Theorem.

Step-by-step explanation:

To determine whether the integrals converge or diverge using the Comparison Theorem, we must compare them with known integrable or non-integrable functions. The Comparison Theorem states that if 0 ≤ f(x) ≤ g(x) for all x ≥ a and ∫ g(x) dx converges, then ∫ f(x) dx must also converge.

(a) For the integral ∫ (x/(x³ + 1)) dx, we can notice that for x > 1, x³ + 1 > x³ and thus x/(x³ + 1) < x/x³ = 1/x². Since the integral of 1/x² from 1 to infinity converges, the given integral also converges.

(b) For the integral ∫ (1/(x ln x)) dx, let us consider the interval [e,∞), for which ln x ≥ 1. Thus, 1/(x ln x) ≤ 1/x, and because the integral of 1/x from e to infinity diverges, the given integral also diverges.