Calculus

What is f(x) if f'(x) = 2x/x^2-1 and f(2) = 0?

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asked by Philip
  1. If

    f'(x) = 2x/x^2-1

    mean

    f'(x) = 2 x / ( x² -1 )

    then

    f (x) = ∫ f'(x) dx = ∫ 2 x dx / ( x² -1 )
    ___________
    Remark:

    Substitution

    x² -1 = u

    2 x dx = du
    __________

    f (x) = ∫ du / u = ℓn ( u ) + C

    f (x) = ℓn ( x² -1 ) + C


    f(2) = 0

    f (2) = ℓn ( 2² -1 ) + C = 0

    ℓn ( 4 -1 ) + C = 0

    ℓn ( 3 ) + C = 0

    Subtract ℓn ( 3 ) to both sides

    ℓn ( 3 ) + C - ℓn ( 3 ) = 0 - ℓn ( 3 )

    C = - ℓn ( 3 )


    f (x) = ℓn ( x² -1 ) + C

    f (x) = ℓn ( x² -1 ) - ℓn ( 3 )

    f (x) = ℓn [ ( x² -1 ) / 3 ]
    ________________________________
    Remark:

    ℓn ( a ) - ℓn ( b) = ℓn ( a / b )

    so

    ℓn ( x² -1 ) - ℓn ( 3 ) = ℓn [ ( x² -1 ) / 3 ]
    ________________________________

    ℓn mean natural logarithm ( logarithm to the base e )

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    posted by Bosnian
  2. Thank you, Bosnian brother.
    That is exactly what I put on my test, I was just making sure.

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    posted by Philip

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