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Math Help - A primitive function for simple equation

  1. #1
    Senior Member Twig's Avatar
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    A primitive function for simple equation

    Hi!

    This shouldnt be hard but for some reason itīs giving me problem.

    f(x) = 1 / (x+1)

    How do I find the F(x) for this function?
    Thanks
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  2. #2
    Moo
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    Hello,

    Note that the derivative for (x+1) is 1.

    So you have something like u'(x)/u(x), with u(x)=1+x.

    The antiderivative of u'(x)/u(x) is ln(|u(x)|)
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  3. #3
    Senior Member Twig's Avatar
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    hi

    so basically I get ln ( x +1 ) ?

    thanks
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by Twig View Post
    so basically I get ln ( x +1 ) ?

    thanks
    Yes, that is your integral.

    I wonder why you posted this in the Algebra forum? The only way to solve for such a question is to use Calculus techniques.
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  5. #5
    Senior Member Twig's Avatar
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    hi

    thanks a lot guys

    So if I have f(x) = 1 / (2x + 2)
    I get F(x) = ln ( 2x + 2 ) ?

    Yeah, I should have posted in a different forum, sorry bout that.
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  6. #6
    Moo
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    Not really, because the derivative for 2x+2 is 2, not 1 :-)

    Or else you can see that f(x)=(1/2)*1/(x+1)
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    GAMMA Mathematics
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    Quote Originally Posted by Twig View Post
    thanks a lot guys

    So if I have f(x) = 1 / (2x + 2)
    I get F(x) = ln ( 2x + 2 ) ?

    Yeah, I should have posted in a different forum, sorry bout that.
    Because of the Chain Rule, you must carry out the derivative of what's inside the natural log. In this case, it is 2, so f(x)=\frac{2}{2x+2}. To make this adjustment for the integral of f(x)=\frac{1}{2x+2}, you must divide the integral by 2, so you have an answer of F(x)=\frac{\ln(2x+2)}{2}.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Twig View Post
    thanks a lot guys

    So if I have f(x) = 1 / (2x + 2)
    I get F(x) = ln ( 2x + 2 ) ?

    Yeah, I should have posted in a different forum, sorry bout that.
    I love doing these general cases \int\frac{a}{bx+c}dx=\frac{a}{b}\int\frac{b}{bx+c}  dx now that we have the derivative of the quantity we can just integrate integrate integrate! \frac{a}{b}\ln|bx+c|+C
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