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Partial fractions integrate 1/(x² − 1) when |x| > 1. The result is
A(1/2) tan⁻¹(x² − 1) + C
B(1/2) ln|(x − 1)/(x + 1)| + C
C(1/2) ln|x² − 1| + C
D(1/2) ln|(x + 1)/(x − 1)| + C
Answer & Solution
Correct answer: B. (1/2) ln|(x − 1)/(x + 1)| + C
Partial fractions give 1/(x² − 1) = (1/2)[1/(x − 1) − 1/(x + 1)], which integrates to (1/2)ln|(x − 1)/(x + 1)|.
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