By partial fractions, ∫ dx/(x(x+1)) equals:
Aln |x| + C
Bln |x(x+1)| + C
Cln |x/(x+1)| + C
Darctan x + C
Answer & Solution
Correct answer: C. ln |x/(x+1)| + C
1/(x(x+1)) = 1/x - 1/(x+1).
We have ∫ = ln |x| - ln |x+1| + C.
So ∫ = ln |x/(x+1)| + C.
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