CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    The value of \[\int{\frac{{{x}^{2}}+1}{{{x}^{2}}-1}}dx\] is

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

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

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

    D)  \[\log ({{x}^{2}}-1)+c\]

    Correct Answer: C

    Solution :

    Let \[I=\int{\frac{{{x}^{2}}+1}{{{x}^{2}}-1}}dx\] \[\Rightarrow \]               \[I=\int{\frac{{{x}^{2}}+1-1+1}{{{x}^{2}}-1}}dx\] \[\Rightarrow \]               \[I=\int{\frac{{{x}^{2}}-1}{{{x}^{2}}-1}dx+\int{\frac{2}{{{x}^{2}}-1}}dx}\] \[\Rightarrow \]               \[I=\int{1\,\,dx\,+2\int{\frac{1}{{{x}^{2}}-1}dx}}\] \[\Rightarrow \]               \[I=x+2.\frac{1}{2}\log \left( \frac{x-1}{x+1} \right)+c\] \[\Rightarrow \]               \[I=x+\log \left( \frac{x-1}{x+1} \right)+c\]


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