The Integral That Will Not Close
A student must evaluate the definite integral of 1/(x² − 1) from 0.5 to 2. He tries a direct antiderivative and gets a nonsensical negative area. He then tries the substitution u = x² − 1, but du = 2x dx does not appear in the integrand, so he is stuck. He notices the integrand blows up at x = 1, which lies inside the interval [0.5, 2], and wonders whether the integral even exists or whether he must split it. He also wants a numerical sanity check to confirm whatever answer he obtains.
1. Explain the singularity at x = 1, decide whether the integral converges, and evaluate it correctly.
Read the worked answer
The integrand 1/(x² − 1) = 1/[(x − 1)(x + 1)] is undefined at x = 1, which is interior to [0.5, 2], so this is an improper integral and must be split: ∫₀.₅² = lim_{b→1⁻} ∫₀.₅^b + lim_{a→1⁺} ∫_a².
Partial fractions: 1/(x² − 1) = ½[1/(x − 1) − 1/(x + 1)].
Antiderivative: ½ ln|x − 1| − ½ ln|x + 1| = ½ ln|(x − 1)/(x + 1)|.
Left piece, b → 1⁻: ½ ln|(b − 1)/(b + 1)| − ½ ln|(−0.5)/(1.5)|. As b → 1⁻, (b − 1) → 0⁻ so the log → −∞. The left piece diverges to −∞.
Right piece, a → 1⁺: the log → +∞, diverging to +∞.
The two one-sided pieces diverge with opposite signs, so the integral does not converge (it is not even a Cauchy principal value in the ordinary sense; the improper integral is divergent). There is no finite area. Any finite number obtained by blind substitution is meaningless - that is why the student's direct attempt gave nonsense.
2. Suppose the interval were changed to [2, 3]. Evaluate the integral there and give a rough numerical check.
Read the worked answer
On [2, 3] there is no singularity, so the fundamental theorem applies directly:
∫₂³ 1/(x² − 1) dx = ½[ln|(x − 1)/(x + 1)|]₂³ = ½[ln(2/4) − ln(1/3)] = ½[ln(1/2) + ln 3] = ½ ln(3/2) ≈ ½(0.4055) ≈ 0.203.
Numerical check: the integrand is 1/3 ≈ 0.333 at x = 2, 1/8 = 0.125 at x = 3, and 1/(2.5² − 1) = 1/5.25 ≈ 0.190 at the midpoint. A trapezium estimate is (1/2)(1)(0.333 + 0.125) ≈ 0.229; Simpson's rule gives (1/6)(0.333 + 4(0.190) + 0.125) ≈ (1/6)(1.218) ≈ 0.203. This matches the exact value, confirming the antiderivative. The lesson: locate singularities before integrating, and always cross-check with an independent numerical estimate.