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Math Help - Integration by Partial Fractions

  1. #1
    Member RedBarchetta's Avatar
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    Integration by Partial Fractions

    <br />
\int {\frac{{2x^4 }}<br />
{{x^3  - x^2  + x - 1}}dx} <br />

    This one gets a little bit tricky. So first off I do long division to make the fraction proper:

    <br />
x^3  - x^2  + x - 1\left){\vphantom{1{2x^4 }}}\right.<br />
\!\!\!\!\overline{\,\,\,\vphantom 1{{2x^4 }}}<br />

    Here is my result. You can check my division although i'm not skilled enough in MathType to show all of it yet.

    <br />
\int {\left( {2x + 2 + \frac{2}<br />
{{x^3  - x^2  + x - 1}}} \right)} dx<br />

    Now I need to find out a way to simplify that last term.

    <br />
\frac{2}<br />
{{x^3  - x^2  + x - 1}} = \frac{2}<br />
{{x(x^2  - x + 1) - 1}}<br />

    This is as far as I went. Any help is appreciated. Thanks!
    Last edited by RedBarchetta; October 7th 2008 at 11:29 AM.
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  2. #2
    Moo
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    Hello,

    As for the last one... Look at the polynomial x^3-x^2+x-1. You can see that the sum of the coefficients is 0. So 1 is a root of it.
    Hence you can factor by (x-1)
    Actually, after some calculations, x^3-x^2+x-1=(x-1)(x^2+1)

    And you can do partial fractions here
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  3. #3
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    As for the last one... Look at the polynomial x^3-x^2+x-1. You can see that the sum of the coefficients is 0. So 1 is a root of it.
    Hence you can factor by (x-1)
    Actually, after some calculations, x^3-x^2+x-1=(x-1)(x^2+1)

    And you can do partial fractions here
    Great. The rest is simple. I was trying to come up with some similar but had no luck. I don't quite understand what you mean one being the root of it. Could you further explain this to me?

    thanks Moo once again.
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  4. #4
    Moo
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    Quote Originally Posted by RedBarchetta View Post
    Great. The rest is simple. I was trying to come up with some similar but had no luck. I don't quite understand what you mean one being the root of it. Could you further explain this to me?

    thanks Moo once again.
    A root is a number that annulates a polynomial.

    Let P(x)=x^3-x^2+x-1

    If the sum of the coefficients is 0, you can say that 1 is a root, that is to say P(1)=0 : P(1)=1^3-1^2+1-1=0

    More generally, let a polynomial Q(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0, with the sum of coefficients equal to 0.

    If we let x=1, we'll have Q(1)=a_n+a_{n-1}+\dots+a_1+a_0 and this equals 0 since the sum of all coefficients is 0. Q(1)=0 hence 1 is a root.

    Now, if a is a root of a polynomial Q(x), then Q(x) is a multiple of (x-a).


    It's a bit messy, but it should be all I can say on the subject :s
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by RedBarchetta View Post
    <br />
\int {\frac{{2x^4 }}<br />
{{x^3  - x^2  + x - 1}}dx} <br />
    First x^3-x^2+x-1=x^2(x-1)+(x-1)=(x-1)(x^2+1). Now (I omitted the constant 2, nothing out the extraordinary),

    \begin{aligned}\int\frac{x^4}{x^3-x^2+x-1}&=\int\frac{(x^2+1)(x^2-1)+1}{(x-1)(x^2+1)}\,dx\\&=\int(x+1)\,dx+\frac12\int\frac{x  ^2+1-(x^2-1)}{(x-1)(x^2+1)}\,dx\\&=\int(x+1)\,dx+\frac12\left(\int\  frac{dx}{x-1}\,dx-\int\frac{x+1}{x^2+1}\,dx\right).\end{aligned}

    Finish off from there, gtg now.
    Last edited by Krizalid; October 7th 2008 at 07:11 PM.
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