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JEE Main & Advanced Sample Paper JEE Main Sample Paper-36

  • question_answer
    If\[\int{\frac{({{x}^{2}}-1)dx}{({{x}^{4}}+3{{x}^{2}}+1){{\tan }^{-1}}\left( \frac{{{x}^{2}}+1}{x} \right)}=\log |{{\tan }^{-1}}f(x)|+C}\], then

    A) \[f(x)={{x}^{2}}+1\]                     

    B) \[f(x)=\frac{{{x}^{2}}+1}{2x}\]

    C) \[f(x)=\frac{{{x}^{2}}+1}{x}\]     

    D) \[f(x)=\frac{1}{2}({{x}^{2}}+1)\]

    Correct Answer: C

    Solution :

    \[I=\int{\frac{({{x}^{2}}-1)dx}{({{x}^{4}}+3{{x}^{2}}+1){{\tan }^{-1}}\left( \frac{{{x}^{2}}+1}{x} \right)}}\] (Dividing Num. and Den. by\[{{x}^{2}})\] \[I=\int{\frac{\left( 1-\frac{1}{{{x}^{2}}} \right)dx}{\left( {{x}^{2}}+3+\frac{1}{{{x}^{2}}} \right){{\tan }^{-1}}\left( x+\frac{1}{x} \right)}=\int{\frac{dt}{\left( {{t}^{2}}+1 \right){{\tan }^{-1}}t}}}\]\[\left( where\,\,t=x+\frac{1}{x}\Rightarrow dt=\left( 1-\frac{1}{{{x}^{2}}} \right)dx \right)\] \[=\log |{{\tan }^{-1}}t|+C=\log \left| {{\tan }^{-1}}\left( \frac{{{x}^{2}}+1}{x} \right) \right|+C\]\[\Rightarrow \]\[f(x)=\frac{{{x}^{2}}+1}{x}\]


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