A) \[\log (1+{{e}^{x}})+c\]
B) \[-\log (1+{{e}^{-x}})+c\]
C) \[-\log (1-{{e}^{-x}})+c\]
D) \[\log ({{e}^{-x}}+{{e}^{-2x}})+c\]
Correct Answer: B
Solution :
Let \[I=\int_{{}}^{{}}{\frac{dx}{1+{{e}^{x}}}}\] \[=\int_{{}}^{{}}{\frac{{{e}^{x}}}{{{e}^{x}}(1+{{e}^{x}})}dx}\] Put \[{{e}^{x}}=t\Rightarrow {{e}^{x}}dx=dt\] \[\therefore \] \[I=\int_{{}}^{{}}{\frac{1}{t(t+1)}dt}\] \[=\int_{{}}^{{}}{\frac{1}{t}dt-\int_{{}}^{{}}{\frac{1}{t+1}}}\] \[=\log \,t-\log (t+1)+c\] \[=-\log \left( \frac{t+1}{t} \right)+c\] \[=-\log (1+{{t}^{-1}})+c\] \[=-\log (1+{{e}^{-x}})+c\]
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