A) \[\frac{1}{e}\log \left( \frac{1+e}{2} \right)\]
B) \[\log \left( \frac{1+e}{2} \right)\]
C) \[\frac{1}{e}\log (1+e)\]
D) \[\log \left( \frac{2}{1+e} \right)\]
Correct Answer: A
Solution :
| [a] Let \[I=\int\limits_{0}^{1}{\frac{dx}{{{e}^{x}}+e}=\int\limits_{0}^{1}{\frac{{{e}^{x}}dx}{{{e}^{x}}({{e}^{x}}+e)}}}\] |
| Put \[{{e}^{x}}=t\Rightarrow {{e}^{x}}dx=dt\] |
| \[I=\int\limits_{1}^{e}{\frac{dt}{t(t+e)}=\frac{1}{e}\int\limits_{1}^{e}{\left( \frac{1}{t}-\frac{1}{t+e} \right)}}\] |
| \[=\frac{1}{e}\int\limits_{1}^{e}{\frac{1}{t}dt-\frac{1}{e}\int\limits_{1}^{e}{\frac{1}{t+e}dt}}\] |
| \[=\frac{1}{e}\left[ \log \,\,t \right]_{1}^{e}-\frac{1}{e}\left[ \log (t+e) \right]_{1}^{e}\] |
| \[=\frac{1}{e}\left[ \log \,\,t-\log (t+e) \right]_{1}^{e}\] |
| \[=\frac{1}{e}\left[ \log \left( \frac{t}{t+e} \right) \right]_{1}^{e}=\frac{1}{e}\left[ \log \left( \frac{e}{2e} \right)-\log \left( \frac{1}{1+e} \right) \right]\] |
| \[=\frac{1}{e}\log \left[ \frac{\frac{1}{2}}{\frac{1}{(1+e)}} \right]=\frac{1}{e}\log \left( \frac{1+e}{2} \right)\] |
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