\(\int\displaystyle\frac{e^{2x}-1}{e^{2x}+1}dx =\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\log|e^x+e^{-x}|+C\): \(C\) is an arbitrary constant
Dividing the numerator and the denominator by \(e^x\):
\( \displaystyle \frac{e^{2x}-1}{e^{2x}+1} = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \)
The numerator is now the derivative of the denominator, since \(\frac{d}{dx}(e^x + e^{-x}) = e^x - e^{-x}\). Using \(\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C\):
\( \displaystyle \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx = \log|e^x+e^{-x}| + C \)
NCERT (2026-27 edition):
Integration by Substitution
