alleryStart learning →Dividing first, what is the integral of (x^2 + 1)/(x^2 - 1)?
Ax + (1/2) log|(x-1)/(x+1)| + c
Bx + log|(x - 1)/(x + 1)| + c
Cx + 2 log|x^2 - 1| + c
Dx + log|x^2 - 1| + c
Answer & Solution
Correct answer: B. x + log|(x - 1)/(x + 1)| + c
Divide to get 1 + 2/(x^2-1); the integral of 2/(x^2-1) is log|(x-1)/(x+1)|.
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