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

  • question_answer
    \[\int_{{}}^{{}}{\left[ \log (\log x)+\frac{1}{{{(\log x)}^{2}}} \right]}\ dx=\]

    A)                 \[x\log (\log x)+\frac{x}{\log x}+c\]

    B)                 \[x\log (\log x)-\frac{x}{\log x}+c\]

    C)                 \[x\log (\log x)+\frac{\log x}{x}+c\]

    D)                 \[x\log (\log x)-\frac{\log x}{x}+c\]

    Correct Answer: B

    Solution :

                    \[\int_{{}}^{{}}{\left[ \log (\log x)+\frac{1}{{{(\log x)}^{2}}} \right]\,dx}=\int_{{}}^{{}}{\log (\log x)dx+\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}}}dx\]                     \[=x\log (\log x)-\int_{{}}^{{}}{\frac{x}{x\log x}\,dx+\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}dx}}\]                     \[=x\log (\log x)-\frac{x}{\log x}-\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}dx+\int_{{}}^{{}}{\frac{1}{{{(\log x)}^{2}}}dx}}\]                     \[=x\log (\log x)-\frac{x}{\log x}+c.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner