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    I have -2x+4-3/2x
    Integrated: -x^2 + 4x -3/2
    I am not sure about the -3/2 because -3/2x gives -3x^-1/2. So -1+1 = 0 and anything to the power of 0 = 1. So that results in 3/2. Is this correct though?


    Thank you in advance
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    (Original post by Wolfram Alpha)
    I have -2x+4-3/2x
    Integrated: -x^2 + 4x -3/2
    I am not sure about the -3/2 because -3/2x gives -3x^-1/2. So -1+1 = 0 and anything to the power of 0 = 1. So that results in 3/2. Is this correct though?


    Thank you in advance
    integration of 1/x is lnx .
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    (Original post by tangotangopapa2)
    integration of 1/x is lnx .

    Sorry, what is Inx?
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    (Original post by Wolfram Alpha)
    Sorry, what is Inx?

    Np. lnx is the natural logarithm of x. I guess Wikipedia does a better job of explaining this than me: https://en.wikipedia.org/wiki/Natural_logarithm :P

    This is because the derivative of ln x is 1/x.
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    (Original post by tangotangopapa2)
    Np. lnx is the natural logarithm of x. I guess Wikipedia does a better job of explaining this than me: https://en.wikipedia.org/wiki/Natural_logarithm :P

    This is because the derivative of ln x is 1/x.
    Ahh...this is c2 stuff and I haven't done logs yet. So in order to integrate the function I must do logs first?
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    (Original post by Wolfram Alpha)
    Ahh...this is c2 stuff and I haven't done logs yet. So in order to integrate the function I must do logs first?
    If you are working with only the algebraic problems (and some trigonometric functions involving only the sin and cos) then you should be fine by just noting that the integration of 1/x is ln x. Otherwise, you might wanna study logarithms first.
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    maybe without logs you could bring the 1/2x to the top by calling it 2x^-1 then integrate that
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    (Original post by tangotangopapa2)
    If you are working with only the algebraic problems (and some trigonometric functions involving only the sin and cos) then you should be fine by just noting that the integration of 1/x is ln x. Otherwise, you might wanna study logarithms first.
    Okay, cheers!
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    (Original post by CrystalSalvatore)
    maybe without logs you could bring the 1/2x to the top by calling it 2x^-1 then integrate that
    Ahh, so is -3/2x equivalent to -6x^-1 if I bring the 2x to the top as you suggested? Thanks in advance.
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    (Original post by CrystalSalvatore)
    maybe without logs you could bring the 1/2x to the top by calling it 2x^-1 then integrate that
    No, you can't. Integration on x^n is (x^(n+1))/(n+1) only when n is not equal to -1.
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    (Original post by Wolfram Alpha)
    I have -2x+4-3/2x
    Integrated: -x^2 + 4x -3/2
    I am not sure about the -3/2 because -3/2x gives -3x^-1/2. So -1+1 = 0 and anything to the power of 0 = 1. So that results in 3/2. Is this correct though?


    Thank you in advance
    Do you mean -3/(2x) or (-3/2)x ?
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    (Original post by notnek)
    Do you mean -3/(2x) or (-3/2)x ?
    The former
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    This thread...


    OP, what is your exam board? If they do not cover the natural logarithm and its derivative in C2 then you probably have the wrong question or written it down incorrectly.
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    (Original post by RDKGames)
    This thread...


    OP, what is your exam board? If they do not cover the natural logarithm and its derivative in C2 then you probably have the wrong question or written it down incorrectly.
    Edexcel. I asked my teacher today and they said natural logs = a C4 technique...I don't know why this Q was with my C2 questions!
 
 
 
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