A) \[\frac{-1}{1+{{e}^{x}}}\]
B) \[\frac{{{e}^{x}}}{1+{{e}^{x}}}\]
C) \[\frac{1}{1+{{e}^{x}}}\]
D) None of these
Correct Answer: A
Solution :
\[I=\int{\frac{dx}{(1+{{e}^{x}})\left( 1+\frac{1}{{{e}^{x}}} \right)}=\int{\frac{{{e}^{x}}dx}{{{(1+{{e}^{x}})}^{2}}}}}\] Let \[1+{{e}^{x}}=t\], \[\therefore \,\,{{e}^{x}}\,dx=dt\] \\[I=\int{\frac{dt}{{{t}^{2}}}=\int{{{t}^{-2}}dt=\frac{{{t}^{-1}}}{-1}=\frac{{{(1+{{e}^{x}})}^{-1}}}{-1}}}=\frac{-1}{1+{{e}^{x}}}\].
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