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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Integration by Substitution

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx=}\] [MP PET 1987]

    A)            \[\log (1+{{x}^{2}})+c\]

    B)            \[\log {{e}^{{{\tan }^{-1}}x}}+c\]

    C)            \[{{e}^{{{\tan }^{-1}}x}}+c\]

    D)            \[{{\tan }^{-1}}{{e}^{{{\tan }^{-1}}x}}+c\]

    Correct Answer: C

    Solution :

                       Putting \[t={{\tan }^{-1}}x\Rightarrow dt=\frac{1}{1+{{x}^{2}}}\,dx,\] we get            \[\int_{{}}^{{}}{\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}\,dx}=\int_{{}}^{{}}{{{e}^{t}}dt}\]\[={{e}^{t}}+c={{e}^{{{\tan }^{-1}}x}}+c.\]


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