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Integrate wrt x^2

  1. Dec 1, 2011 #1
    My friends were discussing about this problem (which they made up themselves).
    They were substituting x^2 for y ([itex]x^{2}[/itex]=y) and thus the answer would come to be log(y+25)
    that is log([itex]x^{2}[/itex]+25)

    I don't think this is the case , i guess that we would be differentiating wrt a 2nd degree curve like a parabola in case of this problem .

    Would you people point out whats the real thing.:confused:
  2. jcsd
  3. Dec 1, 2011 #2


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    If the integral is really as you wrote it,

    [tex]\int d(x^2) \frac{1}{x^2+25},[/tex]

    then you are free to set y = x^2, as in this case your differential is d(x^2), and it already contains the x^2 so you are free to make the substitution.

    If the integral were

    [tex]\int dx \frac{1}{x^2+25},[/tex]

    then you can of course still substitute [itex]y = x^2[/itex], but then the differential term is different: [itex]dx \rightarrow dy/(2x) = dy/(\pm 2\sqrt{y})[/itex]. Note that if this were a definite integral were the limits went from some negative value to a positive one, you would have to split the integral into two pieces, one from the negative value to 0 and one from 0 to the positive value, and then make the substitution, as you need to choose a sign for the square root and it's different for x > 0 and x < 0.
  4. Dec 2, 2011 #3


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    In general, if g(x) is a differentiable function, then dg(x)= g'(x)dx.

    d(x^2)= 2xdx so
    [tex]\int\frac{1}{x^2+ 25}d(x^2)= \int\frac{2x}{x^2+ 25}dx[/tex]

    Now, if you let [itex]u= x^2+ 25[/itex], du= 2xdx and the integral becomes
    [tex]\int\frac{du}{u}= ln(u)+ C= ln(x^2+ 25)+ C[/tex]
    just as before.

    That really is the case.
  5. Dec 2, 2011 #4
    Thanks for the replies :smile:. My teacher also did the same substitution process when i asked him about this.

    These integrations and differentiations are done in with respect to the x-coordinate which is a straight line .
    I was wondering if we could carry out these operations with respect to curves aswell like the [itex]x^{2}[/itex].

    Sorry if this is absurd, but i would be glad if i was clarified .
  6. Dec 3, 2011 #5


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    Yes, you integrate a function f(x) with respect to another function g(x). You are, in effect, "changing the scale". For the Riemann integral, such a function, g, would have to be differentiable and, as I said before, [itex]\int f(x)dg(x)= \int f(x)g'(x)dx[/tex]. For the Stieljes integral, g does not have to differentiable, only increasing.
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