CUET · MATHS · PYQ PAPER 2025
\(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x\) =
- A log \(\left|e^x+e^{-x}\right|+C\) : C is an arbitrary constant
- B log \(\left|e^{2 x}+1\right|+C\) : C is an arbitrary constant
- C log \(\left|e^{2 x}-e^{-x}\right|+C\) : Cis an arbitrary constant
- D log \(\left|e^{2 x}-1\right|+C\) : C is an arbitrary constant
Answer & Solution
Correct Answer
(A) log \(\left|e^x+e^{-x}\right|+C\) : C is an arbitrary constant
Step-by-step Solution
Detailed explanation
\( \int \frac{e^{2 x}-1}{e^{2 x}+1} d x = \int \frac{e^x(e^x - e^{-x})}{e^x(e^x + e^{-x})} d x \) \( = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} d x \) \( = \log \left|e^x+e^{-x}\right|+C \)
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