Question

Find proper substitution

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

• A. $1+e^{-x}\to t$
• B. $-e^{-x}dx\to dt$
• C. $-{\int }_0^1\frac{dt}{t}$
• D. $-{\int }_0^{1}ln\left|t\right|$

Answer 1

The correct option is A $1+e^{-x}\to t$

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

$1+e^{-x}=t$

$\Rightarrow {-e}^{-x}dx=dt$

${\int }_0^1\frac{-dt}{t}\therefore 1+e^{-x}=t$