A) \[\log (1+{{e}^{x}})\]
B) \[-\log (1+{{e}^{-x}})\]
C) \[-\log (1-{{e}^{-x}})\]
D) \[\log ({{e}^{-x}}+{{e}^{-2x}})\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{dx}{1+{{e}^{x}}}}=\int_{{}}^{{}}{\frac{{{e}^{-x}}}{1+{{e}^{-x}}}}\,dx\] Put \[1+{{e}^{-x}}=t\] Þ \[{{e}^{-x}}dx=-dt\], then it reduces to \[-\int{\frac{dt}{t}=-\log t=-\log (1+{{e}^{-x}})}\].
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