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Integration involving Roots

  1. Aug 10, 2006 #1
    I am stuck half way in solving this problem (the square root nominator confuses me) :confused: :
    http://img235.imageshack.us/img235/8459/1mi1.jpg [Broken]

    and I cannot get it to match the answer given on the back of the textbook:
    http://img235.imageshack.us/img235/7041/answerop5.jpg [Broken]

    Please teach me how to solve this problem. Thanks. :)
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Aug 10, 2006 #2
    First making a substitution would help, let u = x1/2. Then you can factor the denominator and possibly break the integrand into partial fractions.
  4. Aug 10, 2006 #3
    The right substitution is [tex]u=x^{\frac{1}{6}}[/tex].

    Thus we have:

    \int \frac{x^{1/2}}{1 + x^{1/3}} dx = 6 \int \frac{u^8}{1+u^2} du =
    6 \int (u^6 - u^4 + u^2 -1 + \frac{1}{1+u^2}) du
  5. Aug 10, 2006 #4
    Whoops, you're right about that, I missed that it was a cube root in the denominator.
  6. Aug 10, 2006 #5


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    Homework Helper

    Only to explain a little bit further. Of course, We then cannot use Partial Fraction if there are roots (square roots, or whatever) in our expression. So, the aim is that, we should make all the square roots, or cube roots in the problem disappear. And so, we choose [tex]u = \sqrt[6]{x}[/tex], or [tex]u ^ 6 = x[/tex].
    We have:
    [tex]u ^ 6 = x \Rightarrow 6 u ^ 5 du = dx[/tex]
    So: [tex]\sqrt{x} = \sqrt{u ^ 6} = u ^ 3[/tex], and [tex]\sqrt[3]{x} = \sqrt[3]{u ^ 6} = u ^ 2[/tex].
    All the roots now have disappeared.
    We then can use Partial Fraction, and integrate it. :)
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