CUET · MATHS · PYQ PAPER 2025
\(\int \frac{d x}{e^x+e^{-x}}\) is equal to
- A \(\tan ^{-1}\left(e^x\right)+c\) : c is an arbitrary constant
- B \(\tan ^{-1}\left(e^{-x}\right)+c\) : c is an arbitrary constant
- C \(\log \left(e^x-e^{-x}\right)+c\) : c is an arbitrary constant
- D \(\log \left(e^x+e^{-x}\right)+c\) : c is an arbitrary constant
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(e^x\right)+c\) : c is an arbitrary constant
Step-by-step Solution
Detailed explanation
\( \int \frac{d x}{e^x+e^{-x}} = \int \frac{e^x}{(e^x)^2+1} dx \) Let \( u = e^x \Rightarrow du = e^x dx \). \( = \int \frac{du}{u^2+1} = \tan^{-1}(u) + c = \tan^{-1}(e^x) + c \)
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