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An integral problem

  1. Mar 30, 2007 #1
    i tried to solve these integral by substitution
    in the first integral i substituted x^2 for t

    in the second integral i substituted e^x

    it didnt work

    how to solve these integrals?

    plz help
     
  2. jcsd
  3. Mar 30, 2007 #2
    here is the file

    here is the file
     

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  4. Mar 30, 2007 #3
    mm i think you must use fraction partial ...and in denominator adding +1 and - 1 . to use fraction partial .

    1/(x^4 +1 +1 -1 ) = 1/(x^4 - 1 )+2

    1/[(x^2-1)(x^2+1)+2]

    1/[(x-1)(x+1)((x^2 -1)+2) +2]

    1/[(x-1)(x+1)((x-1)(x+1)+2)+2]

    and you can use this web http://integrals.wolfram.com/index.jsp
     
  5. Mar 30, 2007 #4

    and

    and respect to second probelm

    int[1/(e^x - 1)]

    put 1 = e^x/e^x

    its become int[ e^x/(e^2x - e^x)]

    put u = e^x subtitution

    du = e^x dx

    du = u dx

    so dx = du/u

    int [u du/(u^2 - u ) u]

    and u =! 0 cuz u = e^x and exponantioal never will be zero

    int[du/u(u-1)] its fraction partial
     
  6. Mar 30, 2007 #5

    Gib Z

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    Homework Helper

    First one is not fun at all, try x.users way.

    For your second one, its a simple substitution.

    [tex]u=\sqrt{e^x-1}[/tex]

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

    So [tex]\int \frac{1}{\sqrt{e^x-1}} dx[/tex] becomes

    [tex]\int \frac{1}{u} \frac{2u}{u^2+1} du[/tex].

    u's cancel out, take out the factor of 2.

    [tex]2\int \frac{1}{1+u^2} du[/tex].

    That integral is easy, arctan.

    [tex]\int \frac{1}{\sqrt{e^x-1}} dx = 2\arctan (\sqrt{e^x-1})[/tex]
     
  7. Mar 31, 2007 #6

    VietDao29

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    Homework Helper

    How can you Partial Fraction this?
    [tex]\frac{1}{(x - 1) (x + 1) ((x - 1)(x + 1) + 2) + 2}[/tex]? :confused:

    For the first one, you can try the following:

    [tex]\int \frac{dx}{x ^ 4 + 1} = \frac{1}{2} \int \frac{(x ^ 2 + 1) - (x ^ 2 - 1)}{x ^ 4 + 1} dx = \frac{1}{2} \left( \int \frac{x ^ 2 + 1}{x ^ 4 + 1} dx + \int \frac{x ^ 2 - 1}{x ^ 4 + 1} dx \right) = \frac{1}{2} (I_1 + I_2)[/tex]

    [tex]I_1 = \int \frac{x ^ 2 + 1}{x ^ 4 + 1} dx = \int \frac{1 + \frac{1}{x ^ 2}}{x ^ 2 + \frac{1}{x ^ 2}} dx = \int \frac{1 + \frac{1}{x ^ 2}}{\left( x - \frac{1}{x} \right) ^ 2 + 2} dx[/tex]

    Now, make the u-substitution: [tex]u = x - \frac{1}{x} \Rightarrow du = \left(1 + \frac{1}{x ^ 2} \right) dx[/tex], your integral will become:

    [tex]I_1 = \int \frac{du}{u ^ 2 + (\sqrt{2}) ^ 2} = \frac{1}{\sqrt{2}} \arctan \left( \frac{u}{\sqrt{2}} \right) + C_1 = \frac{1}{\sqrt{2}} \arctan \left( \frac{x + \frac{1}{x}}{\sqrt{2}} \right) + C_1[/tex]

    The second integral can be done by the u-substitution: [tex]u = x + \frac{1}{x}[/tex]

    [tex]I_2 = \int \frac{x ^ 2 - 1}{x ^ 4 + 1} dx = \int \frac{1 - \frac{1}{x ^ 2}}{x ^ 2 + \frac{1}{x ^ 2}} dx = \int \frac{1 - \frac{1}{x ^ 2}}{\left( x + \frac{1}{x} \right) ^ 2 - 2} dx = ...[/tex]
    Can you go from here? :)
     
    Last edited: Mar 31, 2007
  8. Mar 31, 2007 #7
    thank alot!!!
     
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