JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Self Evaluation Test - Integrals

  • question_answer
    The value of \[\int\limits_{0}^{1}{\frac{dx}{{{e}^{x}}+e}}\] is equal to

    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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