\(\displaystyle\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\log_e|e^{2x}+e^{-2x}|+C\), where C is an arbitrary constant.
The numerator is, up to a factor 2, the derivative of the denominator. Let
\(u = e^{2x} + e^{-2x}\)
\(du = 2\left(e^{2x} - e^{-2x}\right)dx\)
Then
\( \displaystyle \int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx = \frac{1}{2}\int\frac{du}{u} \)
\( \displaystyle = \frac{1}{2}\log_e|u| + C \)
\( \displaystyle = \frac{1}{2}\log_e|e^{2x}+e^{-2x}| + C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{2}\log_e|e^{2x}-e^{-2x}|+C\), where C is an arbitrary constant. — Incorrect. Its derivative is \(\displaystyle\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\), the integrand turned upside down.
Option (2) \(\log_e|e^{2x}-e^{-2x}|+C\), where C is an arbitrary constant. — Incorrect. Its derivative is \(\displaystyle\frac{2(e^{2x}+e^{-2x})}{e^{2x}-e^{-2x}}\).
Option (4) \(\log_e|e^{2x}+e^{-2x}|+C\), where C is an arbitrary constant. — Incorrect. Its derivative is twice the integrand: the factor \(\frac{1}{2}\) from \(du = 2(e^{2x} - e^{-2x})\,dx\) is missing.
NCERT (2026-27 edition):
Integration by Substitution
