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Math Help - [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:

    \int\frac{dx}{x(1+x^n)}
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  2. #2
    Senior Member Peritus's Avatar
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    \int {\frac{1}<br />
{{x\left( {1 + x^n } \right)}}dx} <br />


    \begin{gathered}<br />
  x = \left( {\tan u} \right)^{\frac{2}<br />
{n}}  \hfill \\<br />
  dx = \frac{2}<br />
{n}\left( {\tan u} \right)^{\frac{2}<br />
{n} - 1} \frac{1}<br />
{{\cos ^2 u}} \hfill \\ <br />
\end{gathered}

    <br />
\int {\frac{{\frac{2}<br />
{n}\left( {\tan u} \right)^{\frac{2}<br />
{n} - 1} \frac{1}<br />
{{\cos ^2 u}}}}<br />
{{\left( {\tan u} \right)^{\frac{2}<br />
{n}} \left( {1 + \tan ^2 u} \right)}}} du<br />

    <br />
\int {\frac{2}<br />
{n}\cot udu = \frac{2}<br />
{n}\ln \left| {\sin u} \right| + C} <br />

    now backsubstitute:

    <br />
{\frac{2}<br />
{n}\ln \left| {\sin a\tan x^{\frac{n}<br />
{2}} } \right| + C}<br />

    simplify:

    <br />
\frac{1}<br />
{n}\ln \left| {\frac{{x^n }}<br />
{{1 + x^n }}} \right| + C<br />
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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
    \int\frac{dx}{x(1+x^n)}
    \int {\frac{1}<br />
{{x\left( {1 + x^n } \right)}}\,dx}  = \int {\frac{{\left( {1 + x^n } \right) - x^n }}<br />
{{x\left( {1 + x^n } \right)}}\,dx}  = \int {\frac{1}<br />
{x}\,dx}  - \frac{1}<br />
{n}\int {\frac{{\left( {1 + x^n } \right)'}}<br />
{{1 + x^n }}\,dx} .

    Hence \int {\frac{1}<br />
{{x\left( {1 + x^n } \right)}}\,dx}  = \ln \left| x \right| - \frac{1}<br />
{n}\ln \left( {1 + x^n } \right) + k.
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