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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    If \[\int{\frac{(2{{x}^{2}}+1)\,\,dx}{({{x}^{2}}-4)\,\,({{x}^{2}}-1)}=\log \left[ {{\left( \frac{x+1}{x-1} \right)}^{a}}\,\,{{\left( \frac{x-2}{x+2} \right)}^{b}} \right]}+C,\] then the values of a and b are respectively [Roorkee 2000]

    A) 1/2, ¾

    B) -1, 3/2

    C) 1, 3/2

    D) -1/2, ¾

    Correct Answer: A

    Solution :

    • \[I=\int{\frac{(2{{x}^{2}}+1)}{({{x}^{2}}-4)({{x}^{2}}-1)}dx}\]           
    • \[\frac{2{{x}^{2}}+1}{({{x}^{2}}-4)({{x}^{2}}-1)}=\frac{3}{({{x}^{2}}-4)}-\frac{1}{{{x}^{2}}-1}\]           
    • \[\therefore I=\int{\left[ \frac{3}{({{x}^{2}}-4)}-\frac{1}{{{x}^{2}}-1} \right]}dx\]                  
    • \[=\frac{3}{2\times 2}\log \left| \frac{x-2}{x+2} \right|-\frac{1}{2}\log \left| \frac{x-1}{x+1} \right|+c\]             
    • \[=\frac{3}{4}\log \left| \frac{x-2}{x+2} \right|+\log {{\left| \frac{x+1}{x-1} \right|}^{1/2}}+c\]                 
    • \[=\log {{\left| \frac{x-2}{x+2} \right|}^{3/4}}+\log {{\left| \frac{x+1}{x-1} \right|}^{1/2}}+c\]                      
    • \[=\log \left[ {{\left( \frac{x+1}{x-1} \right)}^{1/2}}{{\left( \frac{x-2}{x+2} \right)}^{3/4}} \right]+c\].


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