BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{x({{x}^{n}}+1)}}\]is equal to:

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

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

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

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

    Correct Answer: C

    Solution :

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


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