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JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If \[\int{\frac{{{x}^{2}}-x+1}{{{x}^{2}}+1}e{{\cot }^{-1}}}x\operatorname{d}x=\operatorname{A}(x){{\operatorname{e}}^{{{\cot }^{-1}}}}x+C,\] then A\[\left( x \right)\] is equal to :     JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[-x\]                                     

    B)  \[x\]

    C)  \[\sqrt{1-x}\]                       

    D) \[\sqrt{1+x}\]

    Correct Answer: B

    Solution :

     Let \[I=\int{\frac{{{x}^{2}}-x+1}{{{x}^{2}}+1}}.{{e}^{{{\cot }^{-1}}x}}dx\] Put \[x=\cot \,t\Rightarrow -\cos e{{c}^{2}}t\,dt=dx\] Now, \[1+{{\cot }^{2}}t=\cos e{{c}^{2}}t\] \[\therefore \]  \[I=\int{\frac{{{e}^{t}}({{\cot }^{2}}t-\cot \,t+1)}{(1+{{\cot }^{2}}t)}(-\cos e{{c}^{2}}t)}dt\] \[=-\int{{{e}^{t}}(\cos e{{c}^{2}}t-\cot \,t)dt}\] \[=\int{{{e}^{t}}(cot\,t-\cos e{{c}^{2}}t)}\,dt\] \[={{e}^{t}}\cot \,t+C\] \[={{e}^{{{\cot }^{-1}}x}}(x)+C\equiv A(x).{{e}^{{{\cot }^{-1}}x}}+C\]                                 \[\Rightarrow \]\[A(x)=x\]


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