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JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    Let\[I=\int_{0}^{1}{\frac{{{e}^{x}}}{x+1}dx}\], then the value of the integral    \[\int_{0}^{1}{\frac{x{{e}^{{{x}^{2}}}}}{{{x}^{2}}+1}dx}\]is

    A) \[{{I}^{2}}\]                                       

    B) \[\frac{1}{2}I\]

    C) \[2I\]                                    

    D) \[\frac{1}{2}{{I}^{2}}\]

    Correct Answer: B

    Solution :

    Let\[{{I}_{1}}=\int_{0}^{1}{\frac{x{{e}^{{{x}^{2}}}}}{{{x}^{2}}+1}dx}\], putting\[{{x}^{2}}=t\] \[\Rightarrow \]               \[2x\,\,dx=dt\], we get \[\therefore \] \[{{I}_{1}}=\frac{1}{2}\int_{0}^{1}{\frac{{{e}^{t}}dt}{t+1}}=\frac{1}{2}\int_{0}^{1}{\frac{{{e}^{x}}dx}{x+1}=\frac{1}{2}I}\]                                 \[\left[ \because \,\,I=\int_{0}^{1}{\frac{{{e}^{x}}dx}{x+1}(given)} \right]\]


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