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Homework Help: Integral with substitution

  1. Mar 22, 2006 #1

    By using a suitable substituition, find

    I haven't encountered this specific type of question before, so I went to use the obvious substitution
    [tex]u^2=1+x^n[/tex], getting:

    [tex]2u=n x^{n-1} \frac{dx}{du}\Leftrightarrow \frac{dx}{du}=\frac{2u}{n} x^{1-n}[/tex]

    [tex]=\int{\frac{2u}{n x^{1-n+1} \sqrt{u^2}}du[/tex]
    [tex]=\frac{2}{n}\left(\frac{1}{3}u^3 - u\right)+c[/tex]
    [tex]=\frac{2}{n}\left[\frac{1}{3}\left(1+x^n\right)^{\frac{3}{2}}-\sqrt{1+x^n}\right] +c[/tex]

    Is it correct?
  2. jcsd
  3. Mar 22, 2006 #2
    [tex]=\int{\frac{2u}{n x^{1-n+1} \sqrt{u^2}}du[/tex]

    This is the first error I found.

    x^(1-n) would be correct if it were in the numerator.
  4. Mar 22, 2006 #3


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    Homework Helper

    I've always found it a bad idea to 'mix' your variables while integrating (the original with the substitution). I think it's better to find expressions for your x's in function of the new variable and then replace all at once.

    Letting [itex]u^2 = 1 + x^n [/itex], we have that:

    x^n = u^2 - 1 \Leftrightarrow x = \left( {u^2 - 1} \right)^{1/n} \Leftrightarrow dx = \frac{{2u}}{n}\left( {u^2 - 1} \right)^{\left( {1 - n} \right)/n} du

    Then substituting

    \int {\frac{1}{{x\sqrt {1 + x^n } }}dx} \to \int {\frac{{2u}}{n}\frac{{\left( {u^2 - 1} \right)^{\left( {1 - n} \right)/n} }}{{\left( {u^2 - 1} \right)^{1/n} u}}du} = \frac{2}{n}\int {\frac{1}{{u^2 - 1}}du}

    As you can see, this simplifies nicely to a classical integral.

    I'll leave integration and substituting back for you :smile:
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