JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    The integral\[\int_{{}}^{{}}{\frac{2{{x}^{12}}+5{{x}^{9}}}{{{x}^{5}}+{{x}^{3}}+1{{)}^{3}}}}dx\]is equal to :-

    A) \[\frac{-{{x}^{10}}}{2{{{{(}^{5}}+{{x}^{3}}+1)}^{2}}}+C\]

    B) \[\frac{-{{x}^{5}}}{{{{{(}^{5}}+{{x}^{3}}+1)}^{2}}}+C\]

    C) \[\frac{{{x}^{10}}}{2{{({{x}^{5}}+{{x}^{3}}+1)}^{2}}}+C\]

    D) \[\frac{{{x}^{5}}}{2{{({{x}^{5}}+{{x}^{3}}+1)}^{2}}}+C\]

    Correct Answer: C

    Solution :

    \[\div \]by \[{{x}^{15}}\]in Nr & Dr\[\int_{{}}^{{}}{\frac{\left( \frac{2}{{{x}^{3}}}+\frac{5}{{{x}^{6}}} \right)dx}{{{\left( 1+\frac{2}{{{x}^{2}}}+\frac{1}{{{x}^{5}}} \right)}^{3}}}}\] Let \[1+\frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{5}}}=t\Rightarrow dt=-\left( \frac{2}{{{x}^{3}}}+\frac{5}{{{x}^{6}}} \right)dx\] \[\int_{{}}^{{}}{\frac{-dt}{{{t}^{3}}}=\frac{1}{2{{t}^{2}}}+c}\] where C is an arbitrary constant.


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