AP EAMCET · Maths · Indefinite Integration
\(\int\left(1+e^{-x}\right)^{-1} d x=\)
- A \(\log \left(1+e^{-x}\right)+c\)
- B \(\log \left(1+e^x\right)+c\)
- C \(\log \left(1-e^x\right)+c\)
- D \(\log \left(e^x-1\right)+c\)
Answer & Solution
Correct Answer
(B) \(\log \left(1+e^x\right)+c\)
Step-by-step Solution
Detailed explanation
\(I=\int\left(1+e^{-x}\right)^{-1} d x=\int \frac{e^x}{e^x+1} d x\) Put \(e^x+1=t \Rightarrow e^x d x=d t\) So, \(I=\int \frac{d t}{t}=\log _e|t|+C=\log _e\left(1+e^x\right)+C\)
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