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Question: Answered & Verified by Expert
$\int\left(1+e^{-x}\right)^{-1} d x=$
MathematicsIndefinite IntegrationAP EAMCETAP EAMCET 2020 (22 Sep Shift 1)
Options:
  • 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 $\log \left(e^x-1\right)+c$
Solution:
2320 Upvotes Verified Answer
The correct answer is: $\log \left(1+e^x\right)+c$
$I=\int\left(1+e^{-x}\right)^{-1} d x=\int \frac{e^x}{e^x+1} d x$
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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