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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    The value of \[\int{\frac{\sqrt{({{x}^{2}}-{{a}^{2}})}}{x}dx}\] will be   [UPSEAT 1999]

    A) \[\sqrt{({{x}^{2}}-{{a}^{2}})}\,-a{{\tan }^{-1}}\left[ \frac{\sqrt{({{x}^{2}}-{{a}^{2}})}}{a} \right]\]

    B) \[\sqrt{({{x}^{2}}-{{a}^{2}})}\,+a{{\tan }^{-1}}\left[ \frac{\sqrt{({{x}^{2}}-{{a}^{2}})}}{a} \right]\]

    C) \[\sqrt{({{x}^{2}}-{{a}^{2}})}\,+{{a}^{2}}{{\tan }^{-1}}[\sqrt{{{x}^{2}}-{{a}^{2}}}]\]

    D) \[{{\tan }^{-1}}x/a+c\]

    Correct Answer: A

    Solution :

    • Let \[\sqrt{({{x}^{2}}-{{a}^{2}})}=t\] Þ \[{{x}^{2}}-{{a}^{2}}={{t}^{2}}\]              
    • Þ \[{{x}^{2}}={{a}^{2}}+{{t}^{2}}\]                   
    • \[\therefore \]\[xdx=tdt\]                   
    • \ \[\int{\frac{\sqrt{({{x}^{2}}-{{a}^{2}})}}{x}dx}=\int{\frac{\sqrt{({{x}^{2}}-{{a}^{2}})}\,x}{{{x}^{2}}}dx}\]           
    • Þ  \[I=\int{\frac{t}{{{a}^{2}}+{{t}^{2}}}tdt}\]\[=\int{\frac{{{t}^{2}}}{{{a}^{2}}+{{t}^{2}}}dt}\]           
    • Þ  \[I=\int{\left( 1-\frac{{{a}^{2}}}{{{a}^{2}}+{{t}^{2}}} \right)\,dt}\]\[=t-{{a}^{2}}\frac{1}{a}{{\tan }^{-1}}\left( \frac{t}{a} \right)\]                
    • Þ  \[I=\sqrt{({{x}^{2}}-{{a}^{2}})}\,-a{{\tan }^{-1}}\left[ \frac{\left\{ \sqrt{({{x}^{2}}-{{a}^{2}})} \right\}}{a} \right]\].


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