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anonymous

  • one year ago

My achilles heel....

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  1. anonymous
    • one year ago
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    |dw:1439113245283:dw|

  2. anonymous
    • one year ago
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    These types of integrals have always been hard, Id appreciate any step by step explanation :)

  3. IrishBoy123
    • one year ago
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    |dw:1439113617615:dw|

  4. IrishBoy123
    • one year ago
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    |dw:1439113662435:dw|

  5. UsukiDoll
    • one year ago
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    ah long division. because the exponent in the denominator is higher than the one int he numerator.

  6. anonymous
    • one year ago
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    Usuki!! :)

  7. UsukiDoll
    • one year ago
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    do you know long division?

  8. anonymous
    • one year ago
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    kind of, I think I do.

  9. UsukiDoll
    • one year ago
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    I have faith in you . you can do it ;)

  10. imqwerty
    • one year ago
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    ^ :D

  11. anonymous
    • one year ago
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    Im afraid faith cant save u or me...help:)?

  12. UsukiDoll
    • one year ago
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    @imqwerty is a good helper. He will guide you ! :D

  13. anonymous
    • one year ago
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    bhaiya help?

  14. imqwerty
    • one year ago
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    •_•)

  15. anonymous
    • one year ago
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    haha dont we all have to do that :)

  16. anonymous
    • one year ago
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    Back to my problem yaar...I got the last exam in calc 2 ever given in sweden, no chance to fail this.

  17. anonymous
    • one year ago
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    Guys....my problem?

  18. IrishBoy123
    • one year ago
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    |dw:1439114866925:dw|

  19. IrishBoy123
    • one year ago
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    finish it off now?

  20. UsukiDoll
    • one year ago
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    don't we have to switch signs for polynomial long division though?

  21. imqwerty
    • one year ago
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    |dw:1439114837536:dw|

  22. UsukiDoll
    • one year ago
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    how the....... O_O @imqwerty what did you do?

  23. imqwerty
    • one year ago
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    :D

  24. anonymous
    • one year ago
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    Guys those methods u showed I dont know them....

  25. imqwerty
    • one year ago
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    |dw:1439115065102:dw|

  26. nincompoop
    • one year ago
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    he used factorization to break the denominator

  27. UsukiDoll
    • one year ago
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    what! everyone should know long division by now. as for those separate integrals.. I see log and arctan

  28. UsukiDoll
    • one year ago
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    I know that we can use factorization on the denominator to get x (x^2+1) . I'm talking about the numerator part. @imqwerty got it uber fast.

  29. anonymous
    • one year ago
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    alright go on imqwerty.

  30. UsukiDoll
    • one year ago
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    I'm 94 SS NOW! XD!

  31. nincompoop
    • one year ago
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    that is not how you properly break the sum of terms with denominator although they yield the same quotient

  32. imqwerty
    • one year ago
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    i broke the equation into 2 parts nd spiltted them nd now we get a cute thing which can be solved easily :)

  33. UsukiDoll
    • one year ago
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    . We know that the exponent in the denominator is huge than the one in the numerator so we need to use long division to fix this problem

  34. anonymous
    • one year ago
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    |dw:1439115395683:dw|

  35. anonymous
    • one year ago
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    but @imqwerty what happened to 2x^2+x+2?

  36. UsukiDoll
    • one year ago
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    that's what I'm wondering too... for the numerator.

  37. imqwerty
    • one year ago
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    did u get how i broke the equation??

  38. UsukiDoll
    • one year ago
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    I got the steps after the equation got broken up

  39. anonymous
    • one year ago
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    could u break the equation, thats the only unclear part. otherwise it was very fast and welldone.

  40. nincompoop
    • one year ago
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    I get how you did it but we do not know how the terms are properly constructed to yield the sum of terms I do not think that is mathematically sound: \(\large \frac{2x^2}{x^3+x}+\frac{x}{x^3+x}+\frac{2}{x^3+x} \rightarrow \frac{2x^3}{x(x^2+1)}...\)

  41. IrishBoy123
    • one year ago
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    |dw:1439115692868:dw|

  42. imqwerty
    • one year ago
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    try takin LCM of those those to terms nd u'll see the original equation appears ok in such questions in which both the numerator and denominator have x our strategy shuld be to break the equation into parts so that all the x terms vanish frm the numerator

  43. IrishBoy123
    • one year ago
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    |dw:1439115748311:dw|

  44. imqwerty
    • one year ago
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    now what i did = (2x^2 + x +2)/(x^3 + x) =2/x + 1/(x^2+1) if u solve this^ then we get ------> [2(x^2+1) + x]/[x(x^2+1)] =(2x^2 + x+2)/(x^3+x) we did nothin wrng we jst altered the equation so as to make it easy to differentiate :)

  45. nincompoop
    • one year ago
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    clever I must say

  46. imqwerty
    • one year ago
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    |dw:1439116006175:dw|

  47. imqwerty
    • one year ago
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    :D

  48. IrishBoy123
    • one year ago
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    |dw:1439116107993:dw|

  49. imqwerty
    • one year ago
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    |dw:1439116153915:dw|

  50. IrishBoy123
    • one year ago
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    |dw:1439116237448:dw|

  51. anonymous
    • one year ago
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    thnx, this wasnt easy I must say.

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