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For |x| < a, the standard integral of 1/(a^2 - x^2) is
A(1/2a) ln|(x - a)/(x + a)| + C
B(1/2a) ln|(a + x)/(a - x)| + C
C(1/a) ln|(a + x)/(a - x)| + C
D(1/a) ln|(x - a)/(x + a)| + C
Answer & Solution
Correct answer: B. (1/2a) ln|(a + x)/(a - x)| + C
Partial fractions give (1/2a)[1/(a - x) + 1/(a + x)], which integrates to (1/2a) ln|(a + x)/(a - x)|.
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