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C3: How to integrate? Watch

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    Afternoon all!
    I know how to integrate simple functions like 3x^2 (add one to the power divide by the new power) but I can't do harder ones.
    I'm stuck on integrating: (2x+1)/(x) and (2x+1)^2/(x)^2.
    Not got a clue where to start
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    (a+b)/c = a/c + b/c
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    (Original post by Get me off the £\?%!^@ computer)
    (a+b)/c = a/c + b/c
    cheers, i got that bit, its the actual integrating i can't do.
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    (Original post by jim666666)
    cheers, i got that bit, its the actual integrating i can't do.
    If you mean you are stuck for the integral of 1/x it is ln x. For the second one you could in this case just expand (2x+1)^2 and divide by x^2 and then integrate.
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    Either expand, or treat it as two functions:
    f(x) = uv , where u=2x+1, and v=2x^-1

    Then f'(x)=u*dv/dx + v*du/dx
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    chain rule? or am i just being stupid saying that? what board are you on?
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    (Original post by luuucyx)
    chain rule? or am i just being stupid saying that? what board are you on?
    Chain rule is for differentiation we want to integrate the functions given
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    (Original post by anshul95)
    Chain rule is for differentiation we want to integrate the functions given
    huumm i knew that :confused: haha
    substitution?
    all ive been shown are realllyyy simple ones of these...
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    (Original post by Dark-Myth)
    Either expand, or treat it as two functions:
    f(x) = uv , where u=2x+1, and v=2x^-1

    Then f'(x)=u*dv/dx + v*du/dx
    i don't even what
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    (Original post by Pheylan)
    i don't even what
    I know it's a bit late...
    but, it's the chain rule.
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    If it's divide, I'd use the quotient rule?
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    (Original post by Dark-Myth)
    I know it's a bit late...
    but, it's the chain rule.
    actually it's incorrect use of the product rule in a situation where the product rule isn't actually supposed to be used
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    (Original post by Pheylan)
    actually it's incorrect use of the product rule in a situation where the product rule isn't actually supposed to be used
    OH ****. It's integration. I didn't realise... my baaaad.
    We don't do integration in C3, so I thought it was differentiation.
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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...
    substution is C4 nice try though
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    (Original post by Get me off the £\?%!^@ computer)
    (a+b)/c = a/c + b/c
    I am confused? There is no integration in C3
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    (Original post by luuucyx)
    chain rule? or am i just being stupid saying that? what board are you on?
    LOL
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    (Original post by Faith01)
    I am confused? There is no integration in C3
    He may be doing a different exam board to what your doing..
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    (Original post by mir3a)
    He may be doing a different exam board to what your doing..
    Oh yeah that makes sense!
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    (Original post by jim666666)
    Afternoon all!
    I know how to integrate simple functions like 3x^2 (add one to the power divide by the new power) but I can't do harder ones.
    I'm stuck on integrating: (2x+1)/(x) and (2x+1)^2/(x)^2.
    Not got a clue where to start
    for the first one do.. let f(x)=2x+1/x, then simplify the expression dividing both 2x and 1 by x. so, f(x)=2+x^-1 then integrate as you would normally.

    For the second one, let f(x)=(2x+1)^2/x^2, expand the numerator, divide by the denominator as we did above, and integrate
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    (Original post by mir3a)
    for the first one do.. let f(x)=2x+1/x, then simplify the expression dividing both 2x and 1 by x. so, f(x)=2+x^-1 then integrate as you would normally.
    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.
 
 
 
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