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Thread: [SOLVED] Antiderivative

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Antiderivative

    I don't really know how to approach this one:

    $\displaystyle \int\frac{dx}{x(1+x^n)}$
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  2. #2
    Senior Member Peritus's Avatar
    Joined
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    $\displaystyle \int {\frac{1}
    {{x\left( {1 + x^n } \right)}}dx}
    $


    $\displaystyle \begin{gathered}
    x = \left( {\tan u} \right)^{\frac{2}
    {n}} \hfill \\
    dx = \frac{2}
    {n}\left( {\tan u} \right)^{\frac{2}
    {n} - 1} \frac{1}
    {{\cos ^2 u}} \hfill \\
    \end{gathered} $

    $\displaystyle
    \int {\frac{{\frac{2}
    {n}\left( {\tan u} \right)^{\frac{2}
    {n} - 1} \frac{1}
    {{\cos ^2 u}}}}
    {{\left( {\tan u} \right)^{\frac{2}
    {n}} \left( {1 + \tan ^2 u} \right)}}} du
    $

    $\displaystyle
    \int {\frac{2}
    {n}\cot udu = \frac{2}
    {n}\ln \left| {\sin u} \right| + C}
    $

    now backsubstitute:

    $\displaystyle
    {\frac{2}
    {n}\ln \left| {\sin a\tan x^{\frac{n}
    {2}} } \right| + C}
    $

    simplify:

    $\displaystyle
    \frac{1}
    {n}\ln \left| {\frac{{x^n }}
    {{1 + x^n }}} \right| + C
    $
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Santiago, Chile
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    Quote Originally Posted by Spec View Post
    $\displaystyle \int\frac{dx}{x(1+x^n)}$
    $\displaystyle \int {\frac{1}
    {{x\left( {1 + x^n } \right)}}\,dx} = \int {\frac{{\left( {1 + x^n } \right) - x^n }}
    {{x\left( {1 + x^n } \right)}}\,dx} = \int {\frac{1}
    {x}\,dx} - \frac{1}
    {n}\int {\frac{{\left( {1 + x^n } \right)'}}
    {{1 + x^n }}\,dx} .$

    Hence $\displaystyle \int {\frac{1}
    {{x\left( {1 + x^n } \right)}}\,dx} = \ln \left| x \right| - \frac{1}
    {n}\ln \left( {1 + x^n } \right) + k.$
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