1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Integral with substitution
  1. Integral substitution? (Replies: 3)

  2. Integral substituting (Replies: 4)