In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.
∫ from 0 to 2 of (dx / (1 - x))

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Identify the integral to be tested for convergence: \(\int_0^2 \frac{dx}{1 - x}\).

Notice that the integrand \(\frac{1}{1 - x}\) has a vertical asymptote at \(x = 1\) because the denominator becomes zero, which lies within the interval of integration \([0, 2]\). This means the integral is an improper integral.

Split the integral at the point of discontinuity to handle it as two improper integrals: \(\int_0^1 \frac{dx}{1 - x} + \int_1^2 \frac{dx}{1 - x}\).

For each integral, rewrite them as limits approaching the point of discontinuity: \(\lim_{t \to 1^-} \int_0^t \frac{dx}{1 - x}\) and \(\lim_{s \to 1^+} \int_s^2 \frac{dx}{1 - x}\).

Evaluate or analyze the behavior of these limits to determine if they converge or diverge. You may use comparison tests or direct integration to assess the convergence of each part.

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Improper Integrals

Improper integrals involve integrals with infinite limits or integrands with discontinuities within the interval. In this problem, the integrand has a vertical asymptote at x = 1, making the integral improper. Understanding how to handle such integrals by taking limits is essential to determine convergence.

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Improper Integrals: Infinite Intervals

Direct Comparison Test

The Direct Comparison Test compares the given integrand to a simpler function whose convergence behavior is known. If the integrand is smaller than a convergent function or larger than a divergent function on the interval, conclusions about convergence or divergence can be drawn.

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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test involves taking the limit of the ratio of the given integrand to a known benchmark function as the variable approaches the problematic point. If the limit is a positive finite number, both integrals share the same convergence behavior, aiding in determining the original integral's convergence.

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