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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Integration by Substitution

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{x}^{2}}+1}{x({{x}^{2}}-1)}\ dx}\] is equal to            [MP PET 1999]

    A)            \[\log \frac{{{x}^{2}}-1}{x}+c\]

    B)            \[-\log \frac{{{x}^{2}}-1}{x}+c\]

    C)            \[\log \frac{x}{{{x}^{2}}+1}+c\]

    D)            \[-\log \frac{x}{{{x}^{2}}+1}+c\]

    Correct Answer: A

    Solution :

                       \[I=\int_{{}}^{{}}{\frac{{{x}^{2}}+1}{x({{x}^{2}}-1)}\,dx}=\int_{{}}^{{}}{\frac{1+\left( \frac{1}{{{x}^{2}}} \right)}{x-\left( \frac{1}{x} \right)}\,dx}\]            Put \[x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{{{x}^{2}}})\,dx=dt\]            \[\therefore \,\,\,I=\int_{{}}^{{}}{\frac{dt}{t}}=\log t+c=\log \frac{{{x}^{2}}-1}{x}+c\].


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