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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    If \[{{I}_{n}}=\int{{{(\log x)}^{n}}\,\,dx},\] then \[{{I}_{n}}+n{{I}_{n-1}}=\] [Karnataka CET 2003]

    A) \[x{{(\log x)}^{n}}\]

    B) \[{{(x\log x)}^{n}}\]

    C) \[{{(\log x)}^{n-1}}\]

    D) \[n{{(\log x)}^{n}}\]

    Correct Answer: A

    Solution :

    • \[{{I}_{n}}=\int{{{(\log x)}^{n}}dx}\]                                     .....(i)                   
    • \[\therefore {{I}_{n-1}}=\int{{{(\log x)}^{n-1}}dx}\]               .....(ii)           
    • Now, \[{{I}_{n}}=\int{{{(\log x)}^{n}}.\,dx}={{(\log x)}^{n}}x-n\int{{{(\log x)}^{n-1}}\frac{1}{x}x\,dx}\]                        
    • \[=x{{(\log x)}^{n}}-n\int{{{(\log x)}^{n-1}}dx}\]                
    • \[{{I}_{n}}=x{{(\log x)}^{n}}-n{{I}_{n-1}}\]; \[\therefore {{I}_{n}}+n\,{{I}_{n-1}}=x{{(\log x)}^{n}}\].


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