\(\int\displaystyle\frac{x^2}{x^2-1}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle x+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|+C\)
The degree of the numerator equals that of the denominator, so the fraction is improper and is divided first:
\( \displaystyle \frac{x^2}{x^2-1} = \frac{(x^2-1)+1}{x^2-1} = 1+\frac{1}{x^2-1} \)
Using \(\int\frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|\) with a = 1:
\( \displaystyle \int\frac{x^2}{x^2-1}\,dx = x+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|+C \)
Why the other options are incorrect
Option (1) \(\displaystyle x-\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|+C\) — Incorrect. The sign of the log term is wrong; it gives \(1-\frac{1}{x^2-1}\).
Option (2) \(\displaystyle x+\log\left|\frac{x-1}{x+1}\right|+C\) — Incorrect. The factor \(\frac{1}{2a} = \frac{1}{2}\) is missing.
Option (3) \(\displaystyle x+\frac{1}{2}\log\left|x^2-1\right|+C\) — Incorrect. \(\frac{1}{2}\log|x^2-1|\) is the integral of \(\frac{x}{x^2-1}\), not of \(\frac{1}{x^2-1}\).