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    Hi

    does anyone know how i can integrate this

    2/8-4x+x^2
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    it would just be -4+2x wouldnt it?
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    Hi!

    Start off with the first argument: If f(x)=2/8x^0 with x^0 = 1, then the integration is as follows: F(x)=(0.25/1)x^(0+1).
    Now, you take f(x)=-4x^1 and attempt to intergrate it in congruence with the example above.
    Finally, it's the same with f(x)=x^2, f(x)=1x^2 respectively.
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    its just simpy 2x/8 -2x^2 + x^3/3 +c
    hannahH93 has differentiated not intergrated...
    dont forget the +c
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    thanks but how did you arrive at that answer
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    (Original post by chris007as)
    Hi

    does anyone know how i can integrate this

    2/8-4x+x^2
    \frac{2}{8}-4x+x^2

    \frac{2}{8-4x+x^2}

    Which of these did you mean?
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    the second part
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    (Original post by chris007as)
    the second part
    In that case, all the answers given so far are wrong - they were all for the 1st equation.
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    (Original post by chris007as)
    the second part
    You'll have to factorise, then split it up into partial fractions, I think.
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    (Original post by + polarity -)
    You'll have to factorise, then split it up into partial fractions, I think.
    Unfortuently this equation can't be factorised.
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    (Original post by starofale)
    Unfortuently this equation can't be factorised.
    Yeah :erm:

    :beard: Complete the square?
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    (Original post by + polarity -)
    Yeah :erm:

    :beard: Complete the square?
    Yes, that's what I was trying, but then it got into trigonometric substitution stuff :mad:
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    (Original post by starofale)
    Yes, that's what I was trying, but then it got into trigonometric substitution stuff :mad:
    Wow! :nothing: What kind of question is this?!
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    Complete the square, do a simple linear sub to get something of the form 1/(a^2+t^2); this is a standard integral that should be in your formula booklet.
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    (Original post by DFranklin)
    Complete the square, do a simple linear sub to get something of the form 1/(a^2+t^2); this is a standard integral that should be in your formula booklet.
    Its in the forumla book?!?

    :eek3: I'd just integrated it myself when I read this
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    is it just (ln(8-4x-x^2))/x-2 + c?
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    (Original post by 1052319933)
    is it just (ln(8-4x-x^2))/x-2 + c?
    No, its to do with a trigonometric function. ln doesn't appear in the answer.

    DFranklin's post says what to do.
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    \dfrac{2}{8-4x+x^2}

    \dfrac{2}{8+(x-2)^2-4}

    \dfrac{2}{4+(x-2)^2}

    \dfrac{1}{2+(\frac{x-2}{4})^2}

    Which, if you look in your formula booklet, integrates to:

    \frac{1}{\sqrt2}tan^{-1}\frac{x-2}{4\sqrt2}
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    (Original post by maxfire)
    Which, if you look in your formula booklet, integrates to:

    \frac{1}{\sqrt2}tan^{-1}\frac{x-2}{4\sqrt2}
    umm...
    the answer is:

    tan^{-1}(\frac{x-2}{2})
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    (Original post by maxfire)
    \dfrac{2}{4+(x-2)^2}

    \dfrac{1}{2+(\frac{x-2}{4})^2}
    I believe your mistake is here.
 
 
 
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