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anonymous

  • 5 years ago

what is the intergral of 1/1 +x^2 dy

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  1. anonymous
    • 5 years ago
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    I think it is( x^3)/(3)+c

  2. amistre64
    • 5 years ago
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    1/1 = 1

  3. amistre64
    • 5 years ago
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    but that would be to obvious to be right :)

  4. amistre64
    • 5 years ago
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    and dy? instead of dx? is that a typo....

  5. amistre64
    • 5 years ago
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    (S) 1/(1+x^2) dx perhaps...

  6. amistre64
    • 5 years ago
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    1/(2x) (S) 2x/(1+x^2) dx ....

  7. amistre64
    • 5 years ago
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    ln(1+x^2) --------- (2x) I wish I knew if I am doing this right :)

  8. anonymous
    • 5 years ago
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    Yes, it's ∫1/(1+x^2)dx. Turns out to be arctan(x), I believe...it's on the tables too.

  9. amistre64
    • 5 years ago
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    DOH!!.... yeah.....

  10. amistre64
    • 5 years ago
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    and I messed p taking that "x" out of it.... should left it in if I was going to do what I was thinking of doing :)

  11. amistre64
    • 5 years ago
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    arctan(x)... 1/sqrt(x^2+1)?? great, i just confused myself lol

  12. anonymous
    • 5 years ago
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    let consider u = 1+x^2 then du = 2x dx and dx = 1/2x du substitute to the integral S u/1/2x du =1/2x.1/2(1+x^2)^2 =1/4x ( 1+ X^2 )^2 would be the answer....

  13. anonymous
    • 5 years ago
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    quantum that answer is correct but how did you work it? bysetting it up 1/2 tan^-1 +1 = 2x^2 + 3/( x^2 + 1)^2 dx

  14. amistre64
    • 5 years ago
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    (S) 1/(1+tan^2(x)) dx ??

  15. amistre64
    • 5 years ago
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    wanna take lazy calc, real calc, and stats over the summer :)

  16. anonymous
    • 5 years ago
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    Okay, here it is:\[\arctan(x)=y\]\[\tan(y)=x\]\[dx/dy = \sec^2(x) = tan^2(x)+1\] We want dy/dx, not dx/dy, so invert it:\[dy/dx = \frac{1}{\tan^2(y)+1}\] and because \[\tan^2(y) = x^2\] then\[dy/dx = \frac{1}{1+x^2.}\] Now we're just integrating it, instead of differentiating the opposite.

  17. anonymous
    • 5 years ago
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    why do(S) 1/(1+tan^2(x)) dx would be???

  18. amistre64
    • 5 years ago
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    yeah :)

  19. anonymous
    • 5 years ago
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    if u work like that...thats brilliant idea quantum,,,but we still need to diferentiate 1 and 1+x^2 to solve the problem...

  20. amistre64
    • 5 years ago
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    suppose I shoulda changed the variable :)

  21. anonymous
    • 5 years ago
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    and i think the answer is 2x/(1+x^2)^2 based on Ur way...

  22. anonymous
    • 5 years ago
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    Sorry, in the third line I should have said sec^2(y) = tan^2(y)+1, sorry if that confused anyone.

  23. amistre64
    • 5 years ago
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    no more confused than usual :)

  24. anonymous
    • 5 years ago
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    We established with that, that the derivative of arctan(x) = 1/(1+x^2). When we integrate the derivative, we're going to get back to arctan(x).

  25. anonymous
    • 5 years ago
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    then tell me the idea that u're supposing tan(y)=x?

  26. amistre64
    • 5 years ago
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    the books call it "trig" substitution... if i recall correctly

  27. anonymous
    • 5 years ago
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    It is trig substitution. :P

  28. anonymous
    • 5 years ago
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    the book of Calculus?...

  29. amistre64
    • 5 years ago
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    yep; all the ones from Newton on down have had it :)

  30. anonymous
    • 5 years ago
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    @tianpradana: If tan^-1(x) = y, then x = tan(y).

  31. amistre64
    • 5 years ago
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    Leibnez even considered it..maybe :)

  32. anonymous
    • 5 years ago
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    oh cheh thanks :-)

  33. amistre64
    • 5 years ago
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    tends to be around, integration by parts, u substitutions..trig subs.... inthat general area :)

  34. anonymous
    • 5 years ago
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    brilliant idea guys...

  35. amistre64
    • 5 years ago
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    its the area I have been avoinding like the plague....

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