MHT CET · Maths · Indefinite Integration
\(\int \frac{1}{e^x+1} \mathrm{~d} x=\)
- A \(x+\log \left(e^x+1\right)+c\), where \(c\) is the constant of integration.
- B \(x-\log \left(e^x+1\right)+c\), where \(c\) is the constant of integration.
- C \(\log \left(\mathrm{e}^x-1\right)+x+\mathrm{c}\), where c is the constant of integration.
- D \(\log \left(\mathrm{e}^x-1\right)-x+\mathrm{c}\), where c is the constant of integration.
Answer & Solution
Correct Answer
(B) \(x-\log \left(e^x+1\right)+c\), where \(c\) is the constant of integration.
Step-by-step Solution
Detailed explanation
\( \int \frac{1}{e^x+1} \mathrm{~d} x = \int \frac{e^x+1-e^x}{e^x+1} \mathrm{~d} x \) \( = \int \left(1 - \frac{e^x}{e^x+1}\right) \mathrm{~d} x \)
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