A) \[\log \left( \frac{1+e}{e} \right)-\frac{1}{e}+1\]
B) \[\log \left( \frac{1+e}{2e} \right)-\frac{1}{e}+1\]
C) \[\log \left( \frac{1+e}{2e} \right)+\frac{1}{e}-1\]
D) None of these
Correct Answer: B
Solution :
Put \[1+{{e}^{-x}}=t\Rightarrow -{{e}^{-x}}dx=dt\], then we have \[I=\int_{2}^{1+\frac{1}{e}}{\frac{(t-1)(-dt)}{t}}=\int_{2}^{1+\frac{1}{e}}{\left( \frac{1}{t}-1 \right)}\,dt\] \[=\left[ {{\log }_{e}}t-t \right]_{2}^{1+\frac{1}{e}}={{\log }_{e}}\left( 1+\frac{1}{e} \right)-\left( 1+\frac{1}{e} \right)-{{\log }_{e}}2+2\] \[={{\log }_{e}}\left( \frac{e+1}{2e} \right)-\frac{1}{e}+1\].
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