∫ (x² + 1)/(x² - 1) dx equals:
Ax + ln |(x-1)/(x+1)| + C
Bx + ln|x| + C
Cx - ln |(x-1)/(x+1)| + C
Dx + C
Answer & Solution
Correct answer: A. x + ln |(x-1)/(x+1)| + C
Divide: (x²+1)/(x²-1) = 1 + 2/(x²-1). Now 2/(x²-1) = 2/((x-1)(x+1)) = 1/(x-1) - 1/(x+1) by partial fractions. So integral = x + ln|x-1| - ln|x+1| + C = x + ln|(x-1)/(x+1)| + C.
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