A) \[\log (1+{{e}^{x}})-x-{{e}^{-x}}+c\]
B) \[\log (1+{{e}^{x}})+x-{{e}^{-x}}+c\]
C) \[\log (1+{{e}^{x}})-x+{{e}^{-x}}+c\]
D) \[\log (1+{{e}^{x}})+x+{{e}^{-x}}+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{{{e}^{-x}}}{1+{{e}^{x}}}\,dx}=\int_{{}}^{{}}{\frac{{{e}^{-x}}{{e}^{-x}}}{{{e}^{-x}}+1}\,dx}\] Put \[{{e}^{-x}}+1=t\Rightarrow -{{e}^{-x}}dx=dt,\] then it reduces to \[-\int_{{}}^{{}}{\frac{(t-1)}{t}\,dt}=\int_{{}}^{{}}{\left( \frac{1}{t}-1 \right)\,dt}\] \[=\log t-t+c=\log ({{e}^{-x}}+1)-({{e}^{-x}}+1)+c\] \[=\log ({{e}^{x}}+1)-x-{{e}^{-x}}-1+c\] \[=\log ({{e}^{x}}+1)-x-{{e}^{-x}}+c\], \[(\because \,\,\,1=\]constant).
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