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anonymous

  • 5 years ago

Find the anti-derivative of

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  1. anonymous
    • 5 years ago
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    \[f(x) = x/(x ^{2} + 1)^{2}\]

  2. anonymous
    • 5 years ago
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    that appears to be an arctan

  3. amistre64
    • 5 years ago
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    x^2 + 1 = 2x we need a 2 up top to ln(x) it

  4. amistre64
    • 5 years ago
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    multiply by 1; or rather 2/2 keep the top but pull out the 1/2

  5. amistre64
    • 5 years ago
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    that might not be good tho :)

  6. amistre64
    • 5 years ago
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    x ---------- (x^2 +1)^2 tan the bottom :) maybe

  7. anonymous
    • 5 years ago
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    *alternate method time* 1/2 * f 2x/(x^4+1) u=x^2 du=2x a=1 arctan(x^2)+C

  8. amistre64
    • 5 years ago
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    tan(t) = x tan^2(t) = x^2 tan^2 + 1 = sec^2 sec^2^2 = sec^4 tan(t) sec^4(t)

  9. anonymous
    • 5 years ago
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    here is the rule I used:

  10. amistre64
    • 5 years ago
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    tan(t) sec^-4(t) maybe :)

  11. amistre64
    • 5 years ago
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    my way you would still have to convert dx to dt...

  12. anonymous
    • 5 years ago
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    \[\int\limits_{ }^{ }1/ (a^2+u^2) = 1/a \arctan(u/a)+C\]

  13. anonymous
    • 5 years ago
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    well, that equation looks terrible...

  14. anonymous
    • 5 years ago
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    wow lol

  15. amistre64
    • 5 years ago
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    tan(t) = x dt sec^2 = dx thats helpful to me i think

  16. amistre64
    • 5 years ago
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    [S] tan(t) sec^-2(t) dt ....maybe

  17. amistre64
    • 5 years ago
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    tan = sin/cos = sin sec-1..... mines just messy lol

  18. anonymous
    • 5 years ago
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    Isn't that straight u substitution? f(x) = x (x^2 + 1) ^ -2 Let u= x^2 + 1 and du = 2x

  19. amistre64
    • 5 years ago
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    [S] 1/2u^2 du ?

  20. anonymous
    • 5 years ago
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    That would give you...\[^{} \int\limits_{}^{} 1/2u ^{2}du\]

  21. amistre64
    • 5 years ago
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    u^-2 = -u^-1

  22. anonymous
    • 5 years ago
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    I mean -2 as the expontent

  23. amistre64
    • 5 years ago
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    -1/2u = -1/2(x^2+1) +C

  24. anonymous
    • 5 years ago
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    \[1/2\int\limits_{}^{}u ^{-2}du\]

  25. anonymous
    • 5 years ago
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    =-1/(2u) + C, then substitue x^2 + 1 back in for u

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