JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    \[\int{\frac{{{e}^{x}}}{(2+{{e}^{x}})+({{e}^{x}}+1)}dx}\]is equal to

    A) \[\log \left( \frac{{{e}^{x}}+1}{{{e}^{x}}+2} \right)+c\]   

    B) \[\log \left( \frac{{{e}^{x}}+2}{{{e}^{x}}+1} \right)+c\]

    C) \[\left( \frac{{{e}^{x}}+1}{{{e}^{x}}+2} \right)+c\]                            

    D) \[\left( \frac{{{e}^{x}}+2}{{{e}^{x}}+1} \right)+c\]

    Correct Answer: A

    Solution :

    Let\[I=\int{\frac{{{e}^{x}}}{(2+{{e}^{x}})({{e}^{x}}+1)}}dx\] Put         \[{{e}^{x}}=t\] \[\Rightarrow \]               \[{{e}^{x}}dx=dt\] \[\therefore \]  \[I=\int{\frac{dt}{(2+t)(t+1)}}\]                 \[=\int{\left[ \frac{1}{(1+t)}-\frac{1}{(2+t)} \right]dt}\]                 \[=\log (1+t)-\log (2+t)+c\]                 \[=\log (1+{{e}^{x}})-\log (2+{{e}^{x}})+c\]                 \[=\log \left( \frac{1+{{e}^{x}}}{2+{{e}^{x}}} \right)+c\]


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