JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    The integral\[\int_{{}}^{{}}{\frac{2{{x}^{3}}-1}{{{x}^{4}}+x}}dx\] is equal to: (Here C is a constant of integration)                            [JEE Main Held on 12-4-2019 Morning]

    A) \[{{\log }_{e}}\left| \frac{{{x}^{3}}+1}{x} \right|+C\]

    B) \[\frac{1}{2}{{\log }_{e}}\frac{{{({{x}^{3}}+1)}^{2}}}{|{{x}^{3}}|}+C\]

    C) \[\frac{1}{2}{{\log }_{e}}\frac{|{{x}^{3}}+1|}{{{x}^{2}}}+C\]

    D) \[{{\log }_{e}}\frac{|{{x}^{3}}+1|}{{{x}^{2}}}+C\]

    Correct Answer: A

    Solution :

    \[\int_{{}}^{{}}{\frac{2{{x}^{3}}-1}{{{x}^{4}}+x}}dx\]           \[\int_{{}}^{{}}{\frac{2x-\frac{1}{{{x}^{2}}}}{{{x}^{2}}+\frac{1}{x}}}dx\]           \[{{x}^{2}}+\frac{1}{x}=t\]           \[\left( 2x-\frac{1}{{{x}^{2}}} \right)dx=dt\]           \[\int_{{}}^{{}}{\frac{dt}{t}=\ell n(t)+C}\]           \[=\ell n\left( {{x}^{2}}+\frac{1}{x} \right)+C\]


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