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

  • question_answer
    If \[A=\int\limits_{0}^{1}{\frac{{{e}^{t}}}{t+1}dt,}\] then \[\int\limits_{0}^{1}{{{e}^{t}}\log (1+t)dt}\] in terms of A equals

    A) \[e\log (A)\]

    B) \[\frac{e}{2}-A\]

    C) \[e-l-\frac{A}{2}\]

    D) \[\frac{e}{2}-l-A\]

    Correct Answer: D

    Solution :

    [d] \[\int\limits_{0}^{1}{{{e}^{t}}}\log (1+t)dt=\left[ {{e}^{t}}.\frac{1}{1+t} \right]_{0}^{1}-\int\limits_{0}^{1}{\frac{{{e}^{t}}}{1+t}dt}\] \[=\frac{e}{2}-1-A\]


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