\(\int\displaystyle\frac{x}{x^{4}-1}\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{4}\log\left|\frac{x^{2}-1}{x^{2}+1}\right|+C\)
The numerator x is half the derivative of \(x^2\), so substitute
\(t = x^2,\quad dt = 2x\,dx\)
Then \(x^4 = t^2\) and
\( \displaystyle \int\frac{x}{x^{4}-1}\,dx = \frac{1}{2}\int\frac{dt}{t^{2}-1} \)
Using the standard integral \(\int\frac{dt}{t^2-a^2} = \frac{1}{2a}\log\left|\frac{t-a}{t+a}\right|+C\) with a = 1:
\( \displaystyle = \frac{1}{2}\times\frac{1}{2}\log\left|\frac{t-1}{t+1}\right|+C \)
\( \displaystyle = \frac{1}{4}\log\left|\frac{x^{2}-1}{x^{2}+1}\right|+C \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{2}\log\left|\frac{x^{2}-1}{x^{2}+1}\right|+C\) — Incorrect. It drops the factor \(\frac{1}{2a} = \frac{1}{2}\) of the standard integral.
Option (3) \(\displaystyle\frac{1}{4}\log\left|\frac{x^{2}+1}{x^{2}-1}\right|+C\) — Incorrect. The fraction is upside down, which changes the sign of the logarithm; it is the negative of the answer.
Option (4) \(\displaystyle\frac{1}{2}\tan^{-1}\left(x^{2}\right)+C\) — Incorrect. Its derivative is \(\frac{x}{1+x^4}\), which has a plus sign in the denominator; \(\tan^{-1}\) comes from \(t^2+1\), not \(t^2-1\).