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Integrating the partial fractions of 1/(x(x + 1)) gives:
Alog |(x + 1) / x|
Blog |x(x + 1)|
Clog |x| + log |x+1|
Dlog |x / (x + 1)|
Answer & Solution
Correct answer: D. log |x / (x + 1)|
Integrating 1/x - 1/(x + 1) gives log |x| - log |x + 1| = log |x/(x + 1)|.
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