Practicing Success
Practicing Success
CUET Preparation Today
$\int \frac{d x}{e^x+e^{-x}}$ is equal to: |
$\tan ^{-1}\left(e^x\right)+c$, where c is a constant $\tan ^{-1}\left(e^{-x}\right)+c$, where c is a constant $\tan ^{-1}\left(e^x+e^{-x}\right)+c$, where c is a constant $\log _e\left(e^x+e^{-x}\right)+c$, where c is a constant |
$\tan ^{-1}\left(e^x\right)+c$, where c is a constant |
The correct answer is Option (1) → $\tan ^{-1}\left(e^x\right)+c$, where c is a constant |
