Question

${\int }_0^1\frac{e^{-x}}{1+e^{-x}}dx=$

• A. $log\left(\frac{1+e}{e}\right)-\frac{1}{e}+1$
• B. $log\left(\frac{1+e}{2e}\right)-\frac{1}{e}+1$
• C. $log\left(\frac{1+e}{2e}\right)+\frac{1}{e}+1$
• D. None of these

Answer 1

The correct option is B $log\left(\frac{1+e}{2e}\right)-\frac{1}{e}+1$

$Put1+e^{-x}=t\Rightarrow -e^{-x}dx=dt,thenwehave$
$I={\int }_2^{1+\frac{1}{e}}\frac{(t-1)(-dt)}{t}={\int }_2^{1+\frac{1}{e}}(\frac{1}{t}-1)dt$
$=[lo{g}_{e}t-t{]}_2^{1+\frac{1}{e}}=lo{g}_e(1+\frac{1}{e})-(1+\frac{1}{e})-lo{g}_{2}2+2$
$=lo{g}_e\left(\frac{e+1}{2e}\right)-\frac{1}{e}+1$