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Using partial fractions, 1/(x^2 - a^2) integrates to

A(1/a) tan^-1(x/a) + C
B(1/2a) ln|(x - a)/(x + a)| + C
C(1/2a) ln|(x + a)/(x - a)| + C
D(1/2a) ln|(a - x)/(a + x)| + C
Answer & Solution
Correct answer: B. (1/2a) ln|(x - a)/(x + a)| + C
Splitting 1/(x^2 - a^2) into partial fractions gives (1/2a)[1/(x - a) - 1/(x + a)], which integrates to (1/2a) ln|(x - a)/(x + a)|.
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