A) \[{{e}^{-1}}+\log 2\]
B) \[-(e+\log 2)\]
C) \[-\left( \frac{1}{e}+\log 2 \right)\]
D) \[-({{e}^{-1}}+\log 3)\]
E) \[-(e+\log 3)\]
Correct Answer: C
Solution :
Let \[I=\int_{0}^{1}{\frac{{{e}^{-x}}dx}{1+{{e}^{x}}}}\] \[=\int_{0}^{1}{\frac{1}{{{e}^{x}}(1+{{e}^{x}})}}dx\] Let \[{{e}^{x}}=t\] \[{{e}^{x}}dx=dt\] \[\therefore \] \[I=\int_{1}^{e}{\frac{1}{{{t}^{2}}(1+t)}}dt\] \[=\int_{1}^{e}{\left( -\frac{t-1}{{{t}^{2}}}+\frac{1}{1+t} \right)}dt\] \[=\int_{1}^{e}{\left( \frac{1}{t+1}-\frac{1}{t}+\frac{1}{{{t}^{2}}} \right)}dt\] \[=\left[ \log (t+1)-\log t-\frac{1}{t} \right]_{1}^{e}\] \[=\log (1+e)-1-\frac{1}{e}-\log 2+1\] \[\therefore \]\[\int_{0}^{1}{\frac{{{e}^{-x}}}{1+{{e}^{x}}}}dx=\log (1+e)-\frac{1}{e}-\log 2\] But \[\int_{0}^{1}{\frac{{{e}^{-x}}dx}{1+{{e}^{x}}}}=\log (1+e)+k\] \[\therefore \] \[k=-\left( \frac{1}{e}+\log 2 \right)\]
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