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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    The value of \[\int{\frac{dx}{3-2x-{{x}^{2}}}}\] will be         [UPSEAT 1999]

    A) \[\frac{1}{4}\log \,\left( \frac{3+x}{1-x} \right)\]

    B) \[\frac{1}{3}\log \,\left( \frac{3+x}{1-x} \right)\]

    C) \[\frac{1}{2}\log \,\left( \frac{3+x}{1-x} \right)\]

    D) \[\log \,\left( \frac{1-x}{3+x} \right)\]

    Correct Answer: A

    Solution :

    • \[\int{\frac{dx}{3-2x-{{x}^{2}}}=\int{\frac{dx}{4-({{x}^{2}}+2x+1)}}}\]                             
    • \[=\int{\frac{dx}{4-{{(x+1)}^{2}}}}\]\[=\int{\frac{dt}{{{(2)}^{2}}-{{t}^{2}}}}\]           
    • Where \[x+1=t,\,\,\,\,\,\therefore dx=dt\]                
    • \[\therefore I=\frac{1}{2.2}\log \left( \frac{2+t}{2-t} \right)\]\[=\frac{1}{4}\log \left( \frac{2+x+1}{2-x-1} \right)\]\[=\frac{1}{4}\log \left( \frac{3+x}{1-x} \right).\]


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