Put 1+e^ x=t ⇒ e^ xdx=dt, then we have I=∫_2^1+1/e(t 1)( dt)/t=∫_2^1+1/e ( 1/t 1 )dt =[log_e t t]_2^1+1/e=log_e ( 1+1/e ) ( 1+1/e ) log_2 2+2 =log_e ( e+ ...

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Property 1

∫01 e-x/1+e-x...

Question

$\int_{0}^{1} \frac{e^{- x}}{1 + e^{- x}} d x =$

l o g (\frac{1 + e}{e}) - \frac{1}{e} + 1

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l o g (\frac{1 + e}{2 e}) - \frac{1}{e} + 1

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l o g (\frac{1 + e}{2 e}) + \frac{1}{e} + 1

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None of these

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Solution

The correct option is B $l o g (\frac{1 + e}{2 e}) - \frac{1}{e} + 1$
$P u t 1 + e^{- x} = t \Rightarrow - e^{- x} d x = d t, t h e n w e h a v e$
$I = \int_{2}^{1 + \frac{1}{e}} \frac{(t - 1) (- d t)}{t} = \int_{2}^{1 + \frac{1}{e}} (\frac{1}{t} - 1) d t$
$= [l o g_{e} t - t]_{2}^{1 + \frac{1}{e}} = l o g_{e} (1 + \frac{1}{e}) - (1 + \frac{1}{e}) - l o g_{2} 2 + 2$
$= l o g_{e} (\frac{e + 1}{2 e}) - \frac{1}{e} + 1$

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