Integral Calculus Question 209
Question: The value of $ \int\limits_0^{1}{\frac{dx}{e^{x}+e}} $ is equal to
Options:
A) $ \frac{1}{e}\log ( \frac{1+e}{2} ) $
B) $ \log ( \frac{1+e}{2} ) $
C) $ \frac{1}{e}\log (1+e) $
D) $ \log ( \frac{2}{1+e} ) $
Show Answer
Answer:
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}{( \frac{1}{t}-\frac{1}{t+e} )}} $ $ =\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}[ \log t ]_1^{e}-\frac{1}{e}[ \log (t+e) ]_1^{e} $ $ =\frac{1}{e}[ \log t-\log (t+e) ]_1^{e} $ $ =\frac{1}{e}[ \log ( \frac{t}{t+e} ) ]_1^{e}=\frac{1}{e}[ \log ( \frac{e}{2e} )-\log ( \frac{1}{1+e} ) ] $ $ =\frac{1}{e}\log [ \frac{\frac{1}{2}}{\frac{1}{(1+e)}} ]=\frac{1}{e}\log ( \frac{1+e}{2} ) $
BETA
