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anonymous

  • 5 years ago

complete the following integral or partial substitution with a. ∫(x^5+2x)/(x^6+6x^2+56)^2 dx

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  1. anonymous
    • 5 years ago
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    ∫(x^5+2x)/(x^6+6x^2+56)^2 dx ∫(x^5+2x) dx - ∫(x^6+6x^2+56)^-2 dx

  2. anonymous
    • 5 years ago
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    You can't do this in products, only in additions. suzi20

  3. dumbcow
    • 5 years ago
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    u = x^6 +6x^2 +56 du = 6x^5 +12x dx du = 6(x^5 +2x) dx

  4. anonymous
    • 5 years ago
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    dv = (x^6+6x^2+56)^-2 dx v = -1 (x^6+6x^2+56)^-1 uv - ∫ v du (x^6 +6x^2 +56)(-(x^6+6x^2+56)^-1) - 1/6∫(-(x^6+6x^2+56)^-1) (x^5 +2x) dx hmm... hard

  5. anonymous
    • 5 years ago
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    Suzi please continue the answer

  6. anonymous
    • 5 years ago
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    u have to do partial again

  7. anonymous
    • 5 years ago
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    Let x^6 + 6x^2 + 56 = t Now, du = 6x^5 + 12x dx = 6(x^5 + 2x) dx The integral becomes, \[1/6\int\limits_{}{}dt/t^2\] \[-1/6t\] Putting back the value of t = -1/6(x^6+6x^2+56)

  8. anonymous
    • 5 years ago
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    I could ask you to complete

  9. anonymous
    • 5 years ago
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    that's calculus 2, i remember now, wow amogh cool...........

  10. anonymous
    • 5 years ago
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    amogh can I ask you to write down the answer from the first until the end if you're willing I will be very thankful to you

  11. anonymous
    • 5 years ago
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    he did

  12. anonymous
    • 5 years ago
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    Tell me what you didn't understand, I've written it completely!

  13. anonymous
    • 5 years ago
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    oh I'm sorry, I really do not understand about the calculus lesson, but I want to learn

  14. anonymous
    • 5 years ago
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    Let x^6 + 6x^2 + 56 = t Differentiating on both sides, Now, dt = (6x^5 + 12x)dx = 6(x^5 + 2x) dx Multiplying the integral by 6 and diving by 6, \[1/6\int\limits_{}^{}6(x^5+2x)/(x^6+6x^2+56)^2 dx\] Now 6(x^5+2x) becomes dt. The integral becomes, \[1/6\int\limits_{}^{} dt/t^2\] = −1/6t Putting back the value of t = -1/6(x^6+6x^2+56)

  15. anonymous
    • 5 years ago
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    amogh thanks a lot

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