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Another substitiution question

  1. Oct 22, 2006 #1
    Evaluate the integral

    [tex] \int \frac {1+x}{1+x^2}dx [/tex]

    i let u = 1+x
    du = dx

    so
    [tex]\int \frac {dx}{u^2}du [/tex]

    or

    [tex]\int u^-2du [/tex]

    [tex] \frac {-1}{u} + C[/tex]

    im not sure if this is right, the only other thing I can think of is to let u = (1+x)/(1+x^2) and then use the quotient rule.

    Any feedback is much appreciated.

    Thanks
     
  2. jcsd
  3. Oct 22, 2006 #2

    arildno

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    No, it is not right.
    Split your integral as follows:
    [tex]\int\frac{1+x}{1+x^{2}}dx=\int\frac{dx}{1+x^{2}}+\int\frac{xdx}{1+x^{2}}[/tex]
    maybe that helps.
     
  4. Oct 22, 2006 #3
    Split the first integral in two.

    Btw, u^2 = (1+x)^2 != 1+x^2
     
  5. Oct 22, 2006 #4
    hmm, that makes sense. Would I evaluate them like this then?

    for the first integral

    u = 1 + x
    du = dx

    and for the second

    u = 1 + x^2
    then du/2 = xdx

    so then

    [tex]\int \frac {1+x}{1+x^2}dx [/tex] = [tex]\int \frac {du}{u^2}[/tex] + [tex]\int \frac {du}{u} [/tex]

    [tex]\frac {-1}{u} [/tex] + ln|u| + C

    thanks again.
     
    Last edited: Oct 22, 2006
  6. Oct 22, 2006 #5

    radou

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    The second one is [tex]\frac{1}{2} \int \frac{du}{u}[/tex]. As for the first one, note that [tex]u^2 = (1+x)^2 \neq 1+ x^2[/tex].
     
  7. Oct 22, 2006 #6
    ohhh, I oversee too many details.

    Can I just leave the integral [tex] \int \frac {dx}{1+x^2} [/tex]

    the way it is and just take the antiderivative to be tan^-1(x) + C
     
  8. Oct 22, 2006 #7

    arildno

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    Why should it be problematic to do so?
     
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