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unkabogable

  • 3 years ago

help me to integrate

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  1. unkabogable
    • 3 years ago
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    |dw:1354504731249:dw|

  2. viniterranova
    • 3 years ago
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    We will see.

  3. Minneemse
    • 3 years ago
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    please someone help me with my problem!!!

  4. anonymous
    • 3 years ago
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    i would start with \[\frac{x+2}{x^2+2x+5}+\frac{1}{x^2+2x+5}\] because the first one gives way to a simple \(u\) sub the second one is a pain though

  5. lrbrock
    • 3 years ago
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    \[\int\limits_{a}^{b} (x+3) * (x^2 + 2x +5)^-1 dx\]

  6. hieuvo
    • 3 years ago
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    |dw:1354504176605:dw|

  7. anonymous
    • 3 years ago
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    i guess not that much of a pain complete the square, solution above

  8. unkabogable
    • 3 years ago
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    |dw:1354505437528:dw| u = x+1 ; a= 2 ?

  9. hieuvo
    • 3 years ago
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    the first one, let u be the denominator, then du will be x+2dx, so this is regular integrate(1/u)du

  10. hieuvo
    • 3 years ago
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    i would make the first numerator x+1, the second numerator is 2

  11. unkabogable
    • 3 years ago
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    okay ..

  12. unkabogable
    • 3 years ago
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    from the top pls.?

  13. unkabogable
    • 3 years ago
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    |dw:1354505864451:dw| ?

  14. hieuvo
    • 3 years ago
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    |dw:1354504775593:dw|

  15. hieuvo
    • 3 years ago
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    and the other one just like above but multiplied by 2 :D

  16. unkabogable
    • 3 years ago
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    for the second one? |dw:1354506253527:dw| kinda impossibele @hieuvo

  17. hieuvo
    • 3 years ago
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    |dw:1354505287482:dw|

  18. unkabogable
    • 3 years ago
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    its unclear to me :(

  19. hieuvo
    • 3 years ago
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    |dw:1354505588414:dw|

  20. hieuvo
    • 3 years ago
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    is that what you're confused ?

  21. unkabogable
    • 3 years ago
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    okay i get it now thanks! :)

  22. hieuvo
    • 3 years ago
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    you're welcome

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