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    Why was 2 divided by 2 & why was the same thing applied to 3?

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    (Original post by Adorable98)


    Why was 2 divided by 2 & why was the same thing applied to 3?

    Video:
    https://youtu.be/eufqV0B6SP8?t=14m44s
    What happens when you differentiate ln(2x+1)?
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    (Original post by Gregorius)
    What happens when you differentiate ln(2x+1)?
    1/(2x+1)
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    (Original post by Adorable98)
    1/(2x+1)
    Now try integrating that and see if you get the same result as when you started...
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    (Original post by Adorable98)
    1/(2x+1)
    Uh uh; how does the chain rule go?
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    (Original post by Adorable98)
    1/(2x+1)
    d/dx ln f(x) = f '(x) / f(x)
    Thus d/dx ln(2x+1) = 2/2x+1
    Therefore Integrating 1/2x+1 would give [1/2 ln(2x+1) ] + C

    As you are multiplying by a constant of 2, the integral turns in [2/2 ln(2x+1)] +C
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    (Original post by Gregorius)
    Uh uh; how does the chain rule go?
    y = Int where t= 2x+1
    dy/dt= 1/t

    dt/dx= 2

    so 1/t x 2
    1/(2x+1) x 2
    2/(2x+1)
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    Same reason as before you need to make the top of the fraction f '(x) = 2
    as the differential of u =(2x+1) is 2
    you do this by
    ∫ 1/(2x+1) dx
    = (1/2)∫ 2/(2x+1) dx
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    (Original post by Katiee224)
    Now try integrating that and see if you get the same result as when you started...
    (Original post by Gregorius)
    Uh uh; how does the chain rule go?
    (Original post by XOR_)
    Same reason as before you need to make the top of the fraction f '(x) = 2
    as the differential of u =(2x+1) is 2
    you do this by
    ∫ 1/(2x+1) dx
    = (1/2)∫ 2/(2x+1) dx
    Okay I get it now Thank you!!
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    (Original post by Adorable98)
    y = Int where t= 2x+1
    dy/dt= 1/t

    dt/dx= 2

    so 1/t x 2
    1/(2x+1) x 2
    2/(2x+1)
    This is where the error is
    you don't times by dt/dx
    you times by dx/dt

    so it's
    ∫ 1/t x (1/2) dt
    = (1/2)∫ 1/t dt
    = (1/2)ln(t) + c
    = (1/2)ln(2x+1) + c

    ^ this is without the constant 2 in the original question if there was a constant 2
    it would be
    = (2/2)ln(2x+1) + c
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    (Original post by XOR_)
    This is where the error is
    you don't times by dt/dx
    you times by dx/dt

    so it's
    ∫ 1/t x (1/2) dt
    = (1/2)∫ 1/t dt
    = (1/2)ln(t) + c
    = (1/2)ln(2x+1) + c

    ^ this is without the constant 2 in the original question if there was a constant 2
    it would be
    = (2/2)ln(2x+1) + c
    But am I not supposed to use the chain rule?
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    (Original post by Adorable98)
    But am I not supposed to use the chain rule?
    if you use t as a variable then
    (dy/dt) = (dy/dx)(dx/dt)

    which is how you get ∫ ..... dt
    so
    ∫ 1/t (dx/dt) dt
    where dx/dt = (1/2)

    -> (1/2)∫ 1/t dt
 
 
 
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