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P3 Integration Challenge watch

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    (Original post by imasillynarb)
    Which answer is right?!
    I don't know, I'm the dummie asking the questions.
    I know how to do these now, I forgot all about integration by partial fractions!!
    Bring on the 50/100 in P3!!!
    Got 52 in my mock, I need 50 ums plus 70 something in S2 to get a B overall!!!
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    (Original post by kikzen)
    1,3
    What the hell have you got there, it's like font 0.9.
    Enlarge please
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    (Original post by Nylex)
    Heh, how'd you know about 1/(1 + x^2) being d/dx (arctan x)? If it's Edexcel P3, they wouldn't want that.. wouldn't they want partial fractions or something instead?
    formula book tbh.
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    (Original post by kikzen)
    formula book tbh.
    Oh seen. It's easy to prove d/dx (arctan x) = 1/(1 + x^2) though.
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    (Original post by BigJohn)
    What the hell have you got there, it's like font 0.9.
    Enlarge please
    you can read it

    (plus i would have to type it up again, and i cant be arsed with that)
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    Ill show you the way I did it, tell me where Im going wrong:

    1/(xlnx) = 1/x.1/lnx

    u = 1/lnx

    To differentiate:

    z = 1
    dz/dx = 0
    s= lnx
    ds/dx = 1/x

    [lnx x 0 - 1/x x 1]/(lnx)^2 = -1/[x(lnx)^2]

    du/dx = -1/[x(lnx)^2]

    dv/dx = 1/x
    v = lnx

    By parts uv - {vdu/dx

    = 1/lnx. lnx - { -1lnx/x(lnx)^2

    = 1 + {1/xlnx

    2{f(x) = 1
    {f(x) = 1/2

    ??
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    (Original post by BigJohn)
    I don't know, I'm the dummie asking the questions.
    I know how to do these now, I forgot all about integration by partial fractions!!
    Bring on the 50/100 in P3!!!
    Got 52 in my mock, I need 50 ums plus 70 something in S2 to get a B overall!!!
    i=int of (x+1)^(2e^x) dx
    Let u=x+1, so that du=dx therefore, I=
    u^(2e^(u+1)) dx
    u = e^lnu, so that I=int of e^2lnu(e^(u+1)) dx
    derivative of 2lnue^(u+1)=((2/u)+(2lnu))(e^u+1)
    So I=e^2lnu(e^(u+1)) / (2/u)+(2lnu))(e^u+1)
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    (Original post by imasillynarb)
    Ill show you the way I did it, tell me where Im going wrong:

    1/(xlnx) = 1/x.1/lnx
    If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
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    Actually now I get 0

    I dont think question 2 can be done.
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    (Original post by Nylex)
    If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
    Ahh yeh got it, surprisingly easy, I dont see why it cant be done using parts though?
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    (Original post by imasillynarb)
    Ahh yeh got it, surprisingly easy, I dont see why it cant be done using parts though?
    why didn't anybody listen to me?? :rolleyes:
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    (Original post by mockel)
    why didn't anybody listen to me?? :rolleyes:
    You didnt say how you did it! I just tried to see if I got the same answer as you
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    (Original post by Nylex)
    If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
    what
    i never learnt that :/ i do everything raw.. but it doesnt work.. :/
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    (Original post by Nylex)
    If you had g(x) = 1/(xln x), you should notice that that's in the form f'(x)/f(x) and so the integral of g(x) = ln [f(x)], ie. ln (ln x) + C.
    a proper solution anyone ?

    not just a shortcut, like.
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    (Original post by kikzen)
    a proper solution anyone ?

    not just a shortcut, like.
    INT 1/(xlnx) dx

    u = lnx
    du/dx = 1/x
    dx = x.du

    = INT [1/(xlnx)].x.du

    'x' cancels, to give: INT (1/lnx) du
    But, lnx = u, so INT (1/u) du
    = ln u + C
    = ln (lnx) + C
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    (Original post by kikzen)
    a proper solution anyone ?

    not just a shortcut, like.
    That's a proper solution damn it.
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    (Original post by mockel)
    INT 1/(xlnx) dx

    u = lnx
    du/dx = 1/x
    dx = x.du

    = INT [1/(xlnx)].x.du

    'x' cancels, to give: INT (1/lnx) du
    But, lnx = u, so INT (1/u) du
    = ln u + C
    = ln (lnx) + C
    If you're working in R0+ (the set of all strictly positive real numbers; greatest set on which ln is defined), you need to take ln¦lnx¦ + C, as lnx is negative if x<1. If lnx<0, then lnx(lnx) does not exist.
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    (Original post by mockel)
    INT 1/(xlnx) dx

    u = lnx
    du/dx = 1/x
    dx = x.du

    = INT [1/(xlnx)].x.du

    'x' cancels, to give: INT (1/lnx) du
    But, lnx = u, so INT (1/u) du
    = ln u + C
    = ln (lnx) + C
    genious!

    ive typed it up (in a bigger font for all you geriatrics) so its a bit clearer. thanks again dude! and nick, of course
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    (Original post by kikzen)
    genious!

    ive typed it up (in a bigger font for all you geriatrics) so its a bit clearer. thanks again dude! and nick, of course
    What program do you do that with?
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    x/(4-x^2)

    x^3+4x^2/(x^2+x-2)

    sin^3thetacos^2theta

    x^3/(X^3-x)

    Won't tell you my answers in case they're wrong. :confused:
    Bloody hell, they ain't this hadr in the book.
 
 
 
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