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anonymous

  • 5 years ago

Evaluate aniti-derivative (7x^4+112x^3+63)/x^5+20x^4+45+2 dx

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  1. anonymous
    • 5 years ago
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    take out 7 as common factor from numerator and put denominator =x and diff

  2. amistre64
    • 5 years ago
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    \[\int\limits_{}\frac{7x^4 +112x^3+63}{x^5+20x^4+45(??)+2}dx\]

  3. anonymous
    • 5 years ago
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    I have gotten to 7(x^4 + 16x^3 + 9)/(x^4 +16x^3 + 9)

  4. anonymous
    • 5 years ago
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    the 45 i left out the x should be 45x

  5. anonymous
    • 5 years ago
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    when diff. denominator we get, 5x^4+80x^3 +45 or 5(x^4+16x^3+9)

  6. anonymous
    • 5 years ago
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    that should be the du isn't it???

  7. anonymous
    • 5 years ago
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    (5x^4+80x^3 +45)dx= dt 5(x^4+16x^3+9)dx=dt (x^4+16x^3+9)dx=dt/5

  8. amistre64
    • 5 years ago
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    mutilpy the top and bottom by 5/7 to get the top as the derivative of the bottom

  9. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{\frac{5}{7}7(x^4+16x^3+9)}{\frac{5}{7}(x^5 +20x^4 +45x +2)}dx\]

  10. amistre64
    • 5 years ago
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    pull out the bottom "5/7" and you got a ln(bottom) as a result:\[\frac{7}{5} \ln(x^5+20x^4+45x+2) +C\]

  11. anonymous
    • 5 years ago
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    didn't we need to substitue somewhere??

  12. amistre64
    • 5 years ago
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    I did, really: u = x^5+20x^4+45x+2 du = 5(x^4 +16^3 +9)

  13. amistre64
    • 5 years ago
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    the top there has a "7" clogging the works; so we multiply by a convient form of "1" to solve it. (5/7)*7 = 5 tada.... so (5/7)/(5/7) can be multiplied to the problem without changing its value...only its "shape"

  14. amistre64
    • 5 years ago
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    We turned it into the form:\[\frac{7}{5}\int\limits_{}\frac{du}{u}\]

  15. anonymous
    • 5 years ago
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    \[7\int\limits_{?}^{?} 1/x^5 + 16x^3 + 9 (x^4+4x^3+2)dx\]in my books examples I have a problem that is similar to this one and I have a step in there that I don't see that you have 7

  16. anonymous
    • 5 years ago
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    between the 9 and (x^4 they are not on the same line th (x^4 is next to the fraction

  17. amistre64
    • 5 years ago
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    \[7 \int\limits_{}\frac{\frac{1}{x^5}+16x^3+9}{x^4+4x^3+2}dx\] like this?

  18. amistre64
    • 5 years ago
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    frac{top}{bottom} is how you get the line between them like that

  19. anonymous
    • 5 years ago
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    I guess it is called differentiate 1/x^5+16x^3+9 then next to this fraction is (x^4+4x^3+2)dx

  20. anonymous
    • 5 years ago
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    1frac{top}{bottom}x^5 +16x^3 +9 x^4+4x^3+2dx

  21. anonymous
    • 5 years ago
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    never mind that didn't work

  22. amistre64
    • 5 years ago
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    \[\int\limits_{}\frac{(x^4+4x^3+2)}{x^5+16x^3+9}dx\] is what the amounts to then...

  23. amistre64
    • 5 years ago
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    \[u = x^5 +16x^3 +9; du=5x^4+48x^2\]

  24. anonymous
    • 5 years ago
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    \[7\int\limits_{?}^{?}\] then the equation you have

  25. amistre64
    • 5 years ago
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    the "du" of that problem isnt wroking to our advantage really. w would need the top part to BE the derivative of the bottom. Unless theres a typo in what you gave me, it looks like a different problem altogether

  26. anonymous
    • 5 years ago
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    well I thought I would try this on my own, but I guess I confussed my self even more. I did more than what I needed. The orginal problem what correct just the 45 needed an x (45x)

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