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

  • question_answer
    \[\int{\frac{\{f(x).\phi '(x)-f'(x).\phi (x)\}}{f(x).\phi (x)}}\log \frac{f(x)}{\phi (x)}dx\] is equal to:

    A) \[\log \frac{\phi (x)}{f(x)}+k\]

    B) \[\frac{1}{2}{{\left\{ \log \frac{\phi (x)}{f(x)} \right\}}^{2}}+k\]

    C) \[\frac{\phi (x)}{f(x)}\log \frac{\phi (x)}{f(x)}+k\]

    D) None of these

    Correct Answer: B

    Solution :

    [b] \[I=\int{\frac{f(x)\phi '(x)-f'(x)\phi (x)}{f(x)\phi (x)}\log \left( \frac{\phi (x)}{f(x)} \right)dx}\] Putting \[\log \frac{\phi (x)}{f(x)}=t\] \[\frac{f(x)}{\phi (x)}.\frac{f(x)\phi '(x)-f'(x)\phi (x)}{{{(f(x))}^{2}}}.dx=dt,\] we get \[I=\int{tdt=\frac{1}{2}{{t}^{2}}+k=\frac{1}{2}{{\left( \log \frac{\phi (x)}{f(x)} \right)}^{2}}+k,k\in R}\]


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