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Thread: Integration of a Rational Function

  1. #1
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    Integration of a Rational Function

    I have a problem that i don't understand what the solutions manual did. I am learning as much calculus 2 over the summer as i can.

    $\displaystyle \int \frac {x^3}{x^2 + 1}\,dx$

    Here is the part where i am confused..

    $\displaystyle \int [ {x} - \frac {x}{x^2 + 1}]\,dx $

    Then,

    $\displaystyle \int {x} \,dx - \frac {1}{2} \int \frac {2x}{x^2 +1} \,dx $

    Finally
    $\displaystyle \frac {1}{2} \, {x^2} - \frac {1}{2} ln(x^2 + 1) + c$
    (using long division)

    I am not sure how they got the answer by the steps. The first step is the most confusing. the second step i can see how they got it. Third step is confusing on how they got

    $\displaystyle \frac {1}{2} \, {x^2} $.
    The rest of it i understand.
    Can anyone explain the first step and the third one?
    Last edited by icelated; Aug 10th 2011 at 11:42 AM.
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  2. #2
    Member anonimnystefy's Avatar
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    re: Integration of a Rational Function

    x^3/(x^2+1)=(x^3+x-x)/(x^2+1)=(x^3+x)/(x^2+1)-x/(x^2+1)=x-x/(x^2+1)
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  3. #3
    Super Member General's Avatar
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    re: Integration of a Rational Function

    First step is just a long division

    or it could be done as follows :

    $\displaystyle \displaystyle \frac{x^3}{x^2+1} = x \cdot \frac{x^2}{x^2+1} = x \cdot \frac{x^2+1-1}{x^2+1} = x \cdot \left( \frac{x^2+1}{x^2+1} - \frac{1}{x^2+1} \right)$

    $\displaystyle \displaystyle = x \cdot \left( 1 - \frac{1}{x^2+1} \right) = x - \frac{x}{x^2+1} $

    and $\displaystyle \frac{1}{2} \, x^2$ is just the integral of x.
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  4. #4
    MHF Contributor Siron's Avatar
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    re: Integration of a Rational Function

    In general:
    $\displaystyle \int x^ndx =\frac{x^{n+1}}{n+1}+C$ if $\displaystyle n\neq -1$
    Use this formule to calculate $\displaystyle \int x^2dx$
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  5. #5
    Member anonimnystefy's Avatar
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    re: Integration of a Rational Function

    $\displaystyle \int\ x\ dx=\frac{1}{2}x^2$

    because

    $\displaystyle \frac{d}{dx}\frac{1}{2}x^2=\frac{1}{2}\cdot2x=1 \cdot x=x$
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  6. #6
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    re: Integration of a Rational Function

    Ok, thanks for the reply. Why do you use a negative one here x^2 +1 - 1 ?
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  7. #7
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    re: Integration of a Rational Function

    Thanks everyone.

    $\displaystyle \frac {1}{2} \, {x^2} $.

    is just the anti dirivitive. i can see it now!!!!
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  8. #8
    MHF Contributor Siron's Avatar
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    re: Integration of a Rational Function

    Quote Originally Posted by icelated View Post
    Ok, thanks for the reply. Why do you use a negative one here x^2 +1 - 1 ?
    If you choose General's way, then you've to subtract -1 if you add +1 to not make a mistake. I mean:
    $\displaystyle \frac{x^3}{x^2+1}=x\cdot \frac{x^2}{x^2+1}=x\cdot \frac{x^2+1-1}{x^2+1}$
    Because the original numerator is $\displaystyle x^2$ so if you add +1 then you have to subtract -1 to hold the original numerator $\displaystyle x^2$.
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  9. #9
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    re: Integration of a Rational Function

    Thanks
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