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    Why is1/(1-x)dx
    = -In|1-x|

    I thought the answer would be In|1-x| without the minus sign in front?
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    (Original post by Adorable98)
    Why is1/(1-x)dx
    = -In|1-x|

    I thought the answer would be In|1-x| without the minus sign in front?
    Nah because you have to use chain rule

    If |1-x| > 0 then |1-x| = 1-x and then the derivative of ln|1-x| is the derivative of ln(1-x) which is -1/(1-x) right? Similarly if |1-x| < 0 then |1-x| = x - 1 and then the derivative is 1/(x-1)= -1/(1-x). So in any case you need to multiply by -1 to get 1/(1-x)
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    (Original post by 13 1 20 8 42)
    Nah because you have to use chain rule

    If |1-x| > 0 then |1-x| = 1-x and then the derivative of ln|1-x| is the derivative of ln(1-x) which is -1/(1-x) right? Similarly if |1-x| < 0 then |1-x| = x - 1 and then the derivative is 1/(x-1)= -1/(1-x). So in any case you need to multiply by -1 to get 1/(1-x)
    I still don't get it .. But when it comes to integrating don't we use the reverse chain rule not the chain rule for differentiation.
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    ∫ f'(x)/f(x) dx = ln(x) + c

    therefore to make it represent the form ∫ f'(x)/f(x) dx
    do: -∫ f'(x)/f(x) dx = -∫ -1/(1-x)dx = -ln(1-x) + c

    you could use the variable u (substitution) if you like:
    u=(1-x)
    du/dx = -1
    ∫ (1/u)(-1) du = -∫ (1/u) du = -ln(u) + c = -ln(1-x) + c
    • Thread Starter
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    (Original post by XOR_)
    ∫ f'(x)/f(x) dx = ln(x) + c

    therefore to make it represent the form ∫ f'(x)/f(x) dx
    do: -∫ f'(x)/f(x) dx = -∫ -1/(1-x)dx = -ln(1-x) + c

    you could use the variable u (substitution) if you like:
    u=(1-x)
    du/dx = -1
    ∫ (1/u)(-1) du = -∫ (1/u) du = -ln(u) + c = -ln(1-x) + c
    I see thank you!!
 
 
 
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