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GUJCET · Maths · Integrals
\(\int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\) __________ \(+C\).
- A \(\log \left(e^{2 x}+1\right)-x\)
- B \(\log \left(e^{2 x}-1\right)+x\)
- C \(\log \left(e^{2 x}+1\right)+x\)
- D \(\log \left(e^{2 x}-1\right)-x\)
Answer & Solution
Correct Answer
(A) \(\log \left(e^{2 x}+1\right)-x\)
Step-by-step Solution
Detailed explanation
\( \int \frac{e^{2 x}-1}{e^{2 x}+1} d x = \int \left(1 - \frac{2}{e^{2 x}+1}\right) d x \) \( = x - 2 \int \frac{e^{-2x}}{1+e^{-2x}} d x \)
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