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Show that F(x)=ln(x^2002+1) is primitive of f(x)=2002x^2001/(x^2002+1)?

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ruals | Salutatorian

Posted September 27, 2013 at 5:28 PM via web

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Show that F(x)=ln(x^2002+1) is primitive of f(x)=2002x^2001/(x^2002+1)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 27, 2013 at 5:51 PM (Answer #1)

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You need to remember that differentiating the primitive function `F(x)` yields `f(x)` , hence, you need to test if `F'(x) = f(x)` , such that:

`F'(x) = (ln(x^2002+1))'`

You need to use the chain rule to evaluate the derivative of the function, such that:

`F'(x) = 1/(x^2002+1)*(x^2002+1)'`

`F'(x) = 1/(x^2002+1)*(2002x^(2002-1))`

`F'(x) = (2002x^2001)/(x^2002+1)`

Comparing the expression of derivative `F'(x)` with the equation of the function `f(x)` yields that they coincide.

Hence, checking if the function `F(x)` is the primitive of `f(x)` , yields that `F'(x) = f(x),` thus, the statement holds.

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