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Can you help me to solve this integral?

  1. Apr 17, 2006 #1

    Nea

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    Can you help me to solve this integral?
    [tex]\int {\frac{{x^3 + 1}}{{x(x^3 - 8)}}} dx[/tex]
     
  2. jcsd
  3. Apr 17, 2006 #2
    Partial fractions? x^3 - 8 = x^3 - 2^3 is easy to factor.
     
  4. Apr 17, 2006 #3

    arildno

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    Dearly Missed

    Remember that [itex]8=2^{3}[/itex]
    Hence, we have:
    [itex]x^{3}-8=(x-2)(x^{2}+2x+4)[/itex]
    Use partial fractions decomposition; it might help to note that:
    [tex]x^{2}+2x+4=(x+1)^{2}+3=3((\frac{x+1}{\sqrt{3}})^{2}+1)[/tex]
     
  5. Apr 17, 2006 #4


    Formula for integration by parts by Substitution Rule:
    [tex]\int u dv = uv - \int v du[/tex]

    [tex]u = x^3 + 1 \; \; \; dv = \frac{dx}{x^4-8x}[/tex]

    [tex]du = 3x^2 dx \; \; \; v = \left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right][/tex]

    [tex]\int \left(x^3 + 1\right)\left(\frac{1}{x^4-8x}\right)dx = \left(x^3 + 1\right)\left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right] - \int \left[\frac{\ln \left(x^3 - 8\right)}{24} - \frac{\ln\left(x\right)}{8}\right]\left(3x^2\right)dx[/tex]
     
    Last edited: Apr 17, 2006
  6. Apr 18, 2006 #5

    VietDao29

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    Orion1, do you think that the latter integral look much more complicated than the former one?!?!
     
  7. Apr 18, 2006 #6
    Affirmative
    [tex]\int \left(x^3 + 1\right)\left(\frac{1}{x^4-8x}\right)dx = \frac{1}{8} \left[ \left(x^3 + 1\right)\left[\frac{\ln \left(x^3 - 8\right)}{3} - \ln\left(x\right)\right] - \int \left[ \ln \left(x^3 - 8\right) - 3 \ln\left(x\right)\right]\left(x^2\right)dx\right][/tex]
     
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