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Problem with integral

  1. Oct 8, 2005 #1
    Ok, so I have the integral of (x^2 + 1) / (x^3 + 8) dx

    I use partial fractions and end up with two integrals, the one I cannot solve is:

    1/12 * "integral of" (7x-4) / [(x-1)^2 +3 ] dx
    With u=x-1 I get
    (7u-3) / (u^2+3) du

    But I have no Idea of how to solve it, the answer in the books shows the step up to this, but jumps over what to do with the last part and writes the answer (something involving tan^-1 and ln)

    Please help!
    Thanks
     
  2. jcsd
  3. Oct 8, 2005 #2
    Factor out the 3 in the denominator:

    [tex]\int\frac{7u-3}{u^2+3}\,du=\frac{1}{3}\int\frac{7u-3}{\left(\frac{u}{\sqrt{3}}\right)^2+1}\,du[/tex]

    Then make the subsititution [itex]u=\sqrt{3}\tan{\theta}[/itex].

    Alex
     
  4. Oct 9, 2005 #3
    Oh, thanks!
     
  5. Oct 9, 2005 #4
    Ok I'm very confused about the doing substitution _two_ times, can someone please go threw every step of this?
     
  6. Oct 9, 2005 #5

    arildno

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    Fra børjan av???
     
  7. Oct 9, 2005 #6
    Sorry, nevermind! I think I can figure this out, can someone please just tell me what the integral of Tan[x] equals?
     
  8. Oct 9, 2005 #7

    arildno

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    It is C-ln(|cos(x)|), where C is an integration constant.
     
  9. Oct 9, 2005 #8
    Ok, I see it now. Took some time :P

    Thanks!
     
  10. Oct 9, 2005 #9
    Ok, bah. I don't get it. Can someone please take it from what apmcavoy wrote (start with subsitution of u)
     
  11. Oct 9, 2005 #10

    arildno

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    You have then:
    [tex]du=\frac{\sqrt{3}d\theta}{\cos^{2}\theta}[/tex]
    Since [tex]tan^{2}\theta+1=\frac{1}{\cos^{2}\theta}[/tex]
    we get:
    [tex]\frac{1}{3}\int\frac{7u-3}{(\frac{u}{\sqrt{3}})^{2}+1}du=\frac{1}{3}\int\sqrt{3}(7tan\theta-3)d\theta[/tex]
    All right?
     
  12. Oct 9, 2005 #11
    Yeah, thanks. I made it to the last step, but I dont see what happens with the Cos bit... :P

    (EDIT: The cos bit in the solution of the integral of Tan phi)
     
  13. Oct 9, 2005 #12

    arildno

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    Okay, you'll basically need to rearrange an expression like cos(Atan(y)), which will appear within the logarithm.

    In order to do this, not that by definition of the Atan and tan functions, we have:
    [tex]y=\tan{Atan(y)}=\frac{\sin(Atan(y))}{\cos(Atan(y))}=\frac{\sqrt{1-\cos^{2}(Atan(y))}}{\cos(Atan(y))}(1)[/tex]
    where I've used the identity [itex]\sin^{2}x=1-\cos^{2}x[/itex] for all x.

    Thus, from (1), we get [tex]y^{2}\cos^{2}(Atan(y))=1-\cos^{2}(Atan(y))[/tex]
    by which we have:
    [tex]|\cos(Atan(y))|=\frac{1}{\sqrt{1+y^{2}}}[/tex]

    Thus, we have rewritten the troublesome expression, in that we now have:
    [tex]ln|\cos(Atan(y))|=-\frac{1}{2}ln(y^{2}+1)[/tex]
     
  14. Oct 9, 2005 #13
    Ok, this made it clear. Thank you!
     
  15. Oct 9, 2005 #14

    arildno

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    You're welcome.
     
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