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    (Original post by luuucyx)
    huumm i knew that :confused: haha
    substitution?
    all ive been shown are realllyyy simple ones of these...
    You could use chain rule, but backwards. That would work!!

    1/a x 1/(n+1) x (ax+b)^(n+1)
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    (Original post by sohail.s)
    I think im missing something here, help is appreciated, if you integrate x^-1 its not possible is it? you'd end up dividing by 0 ??

    Bear in mind i'm still in AS year and studying C2 atm.
    sorry my explanation wasn't as clear...integrating 1/x gives you ln x, its a rule u should learn.
    Let me know if you get stuck
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    (Original post by mir3a)
    sorry my explanation wasn't as clear...integrating 1/x gives you ln x, its a rule u should learn.
    Let me know if you get stuck
    ah ok, that makes sence thanks, all this will help once i start C3 next year, thanks
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    (Original post by Eloades11)
    substution is C4 nice try though
    nah ocr its C3...

    (Original post by Freerider101)
    You could use chain rule, but backwards. That would work!!

    1/a x 1/(n+1) x (ax+b)^(n+1)
    i think i love you, only polite person on this thread haha
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    (Original post by luuucyx)
    nah ocr its C3...
    really? which OCR, im on OCR and I have my C3 exam next month, ive been through 6 past papers and never come across any integration by substitution.

    Unless its OCR gateway or 21 century of any other weird OCR thing, in that case ill let you off
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    (Original post by Eloades11)
    really? which OCR, im on OCR and I have my C3 exam next month, ive been through 6 past papers and never come across any integration by substitution.

    Unless its OCR gateway or 21 century of any other weird OCR thing, in that case ill let you off
    hmmm maybe my teachers done it differently? i do c3 and c4 in one block so imnot sure, but i know i do the ocr main one.
    i spose im a girl :/ ha
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    How has this thread stretched to 2 pages...

    (Original post by sohail.s)
    I think im missing something here, help is appreciated, if you integrate x^-1 its not possible is it? you'd end up dividing by 0 ??

    Bear in mind i'm still in AS year and studying C2 atm.
    This is equivalent to showing that the derivative of \ln x is \frac{1}{x} because integration is the reverse of differentiation.
    Note that \frac{d}{dx}[\ln x] = \displaystyle\lim_{h\to 0} \frac{\ln (x+h)-\ln (x)}{h} = \displaystyle\lim_{h\to 0} \frac{\ln (1+\frac{h}{x})}{h} = \displaystyle\lim_{h\to 0} \ln (1 + \frac{h}{x})^{\frac{1}{h}}
    Now if you let \frac{h}{x}=\frac{1}{u} \implies \frac{1}{h} = \frac{u}{x}.
    Therefore we have:
    \frac{d}{dx}[\ln x]=\displaystyle\lim_{u\to \infty} \ln (1+\frac{1}{u})^{\frac{u}{x}} = \displaystyle\lim_{u\to \infty} \frac{1}{x}\ln (1+\frac{1}{u})^{u}.

    Note that e= \displaystyle\lim_{u\to \infty}(1+\frac{1}{u})^{u}
    Therefore \frac{d}{dx}[\ln x] = \frac{1}{x}\ln (e) = \frac{1}{x} as required.
    Therefore \displaystyle\int\frac{1}{x}dx = \ln x + C.
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    (2x + 1)/x
    goes to
    (2x/x) + (1/x)

    then integrate

    to 2x + in [x] + c

    the second one uses the same method....

    (2x+1)^2/x^2

    goes to

    (4x^2/x^2) + (4x/x^2) + (1/x^2)

    then integrating you get

    4x + 4In [x] - 1/x

    simplifies to

    4x + In [x^4] -1/x (although this step is probably unnecessary)
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    (Original post by sete)
    (2x + 1)/x
    goes to
    (2x/x) + (1/x)

    then integrate

    to 2x + in [x] + c



    2x + in

    Is the in key the one next to the iog key?
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    (Original post by Farhan.Hanif93)
    How has this thread stretched to 2 pages...


    This is equivalent to showing that the derivative of \ln x is \frac{1}{x} because integration is the reverse of differentiation.
    Note that \frac{d}{dx}[\ln x] = \displaystyle\lim_{h\to 0} \frac{\ln (x+h)-\ln (x)}{h} = \displaystyle\lim_{h\to 0} \frac{\ln (1+\frac{h}{x})}{h} = \displaystyle\lim_{h\to 0} \ln (1 + \frac{h}{x})^{\frac{1}{h}}
    Now if you let \frac{h}{x}=\frac{1}{u} \implies \frac{1}{h} = \frac{u}{x}.
    Therefore we have:
    \frac{d}{dx}[\ln x]=\displaystyle\lim_{u\to \infty} \ln (1+\frac{1}{u})^{\frac{u}{x}} = \displaystyle\lim_{u\to \infty} \frac{1}{x}\ln (1+\frac{1}{u})^{u}.

    Note that e= \displaystyle\lim_{u\to \infty}(1+\frac{1}{u})^{u}
    Therefore \frac{d}{dx}[\ln x] = \frac{1}{x}\ln (e) = \frac{1}{x} as required.
    Therefore \displaystyle\int\frac{1}{x}dx = \ln x + C.
    ah ok, cheers, i hope i dont have to remember all that :P
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    (Original post by sohail.s)
    ah ok, cheers, i hope i dont have to remember all that :P
    You don't, I was just showing you where it came from.
 
 
 
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