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Math Help - Partial Fraction Integration

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    Partial Fraction Integration

    Q: f(x) \equiv \frac{1}{(x+2)(x^2+4)}. Express f(x) in partial fractions.

    Show that \displaystyle\int^2_0 f(x) \, \mathrm{d}x = \frac{(\pi + 2 \ln 2)}{32}


    I can't seem to get the partial fraction right. If possibe, can someone help me with this question for me. Thanks. .
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Air View Post
    Q: f(x) \equiv \frac{1}{(x+2)(x^2+4)}. Express f(x) in partial fractions.

    Show that \displaystyle\int^2_0 f(x) \, \mathrm{d}x = \frac{(\pi + 2 \ln 2)}{32}


    I can't seem to get the partial fraction right. If possibe, can someone help me with this question for me. Thanks. .
    did you start by doing \frac 1{(x + 2)(x^2 + 4)} = \frac A{x + 2} + \frac {Bx + C}{x^2 + 4} ?
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    Moo
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    Hello,

    \frac{1}{(x+2)(x^2+4)}=\frac{a}{x+2}+\frac{bx+c}{x  ^2+4}

    What did you do for your partial fraction ? (It's always better to point out mistakes than doing the whole thing )
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    I was doing \frac{1}{(x+2)(x^2+4)}=\frac{a}{x+2}+\frac{b}{x^2+  4} hence my mistake.
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    Splitting into partial fractions: \frac{1}{(x+2)\left(x^{2}+4\right)} = \frac{A}{x+2} + \frac{Bx + C}{x^{2} + 4}

    So, multiplying both sides by (x+2)(x^2 + 4) we have:
    1 = A\left(x^{2}+4\right) + (Bx + C)(x + 2)
    1 = Ax^{2} + 4A + Bx^{2} + (2B+C)x + 2C (multiplied it out)
    1 = (A+B)x^{2} + (2B + C)x + 4A + 2C (grouped)

    If we write the left hand side as a quadratic:
    0x^{2} + 0x + 1 = (A+B)x^{2} + (2B + C)x + (4A + 2C)

    We can equate the coefficients to see that:
    A + B = 0
    2B + C = 0
    4A + 2C = 1

    So this should be an easy system to solve for and you should be on your way.

    Edit: Ok a little too late lol xD.
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    Moo
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    Another way of doing it is :

    - firstly, multiply by (x+2), then take x=-2. It'll directly yield A.

    - secondly, take x=0, to directly get C.

    - thirdly...make your own way to find B
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    o_O
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    Quote Originally Posted by Moo View Post
    Another way of doing it is :

    - firstly, multiply by (x+2), then take x=-2. It'll directly yield A.

    - secondly, take x=0, to directly get C.

    - thirdly...make your own way to find B
    Yes ... That's what I meant ...

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    Quote Originally Posted by o_O View Post
    Yes ... That's what I meant ...

    You multiplied all in once by (x+2)(x+4) ^^

    Actually, I was taught this method for things containing more partial fractions, it was really faster
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Air View Post
    I was doing \frac{1}{(x+2)(x^2+4)}=\frac{a}{x+2}+\frac{b}{x^2+  4} hence my mistake.
    yes. with partial fractions, the numerator must be one less degree than the denominator. so if it was \frac 1{(x + 3)(x^3 + 4x + 2)} for instance. your partial fractions would be \frac A{x + 3} + \frac {Bx^2 + Cx + D}{x^3 + 4x + 2}

    there are other rules for partial fractions. like decomposing \frac 1{x^2} becomes \frac Ax + \frac B{x^2} for instance. make sure you look these rules up
    Last edited by Jhevon; May 4th 2008 at 02:41 PM.
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  10. #10
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    there are other rules for partial fractions. like decomposing \frac 1{x^2} becomes \frac Ax + B{x^2} for instance. make sur eyou look these rules up
    Wait a second, is the decompostition of \frac 1{x^2}= \frac Ax + B{x^2}? Or was that a typo for  \frac {A}{x} + \frac{B}{x^2}?

    If it was the first one I failed my test miserably.
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    Quote Originally Posted by Gusbob View Post
    Wait a second, is the decompostition of \frac 1{x^2}= \frac Ax + B{x^2}? Or was that a typo for  \frac {A}{x} + \frac{B}{x^2}?

    If it was the first one I failed my test miserably.
    Don't panic. If the factor in the denominator is x^2 then the corresponding partial fractions are  \frac {A}{x} + \frac{B}{x^2}.
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    Quote Originally Posted by Air View Post

    Q: f(x) \equiv \frac{1}{(x+2)(x^2+4)}. Express f(x) in partial fractions.

    Show that \displaystyle\int^2_0 f(x) \, \mathrm{d}x = \frac{(\pi + 2 \ln 2)}{32}
    \begin{aligned}<br />
   \frac{1}{(x+2)\left( x^{2}+4 \right)}&=\frac{\left( x^{2}+4 \right)-\left( x^{2}-4 \right)}{8(x+2)\left( x^{2}+4 \right)} \\ <br />
 & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x-2}{x^{2}+4} \right\} \\ <br />
 & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x}{x^{2}+4}+\frac{2}{x^{2}+4} \right\}. <br />
\end{aligned}

    Integrate.
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    Quote Originally Posted by Gusbob View Post
    Wait a second, is the decompostition of \frac 1{x^2}= \frac Ax + B{x^2}? Or was that a typo for  \frac {A}{x} + \frac{B}{x^2}?

    If it was the first one I failed my test miserably.
    yes, it was a typo. i just caught it. i fixed it
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    \begin{aligned}<br />
   \frac{1}{(x+2)\left( x^{2}+4 \right)}&=\frac{\left( x^{2}+4 \right)-\left( x^{2}-4 \right)}{8(x+2)\left( x^{2}+4 \right)} \\ <br />
 & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x-2}{x^{2}+4} \right\} \\ <br />
 & =\frac{1}{8}\left\{ \frac{1}{x+2}-\frac{x}{x^{2}+4}+\frac{2}{x^{2}+4} \right\}. <br />
\end{aligned}

    Integrate.
    How do you see these things?!

    i wanted to do algebraic manipulation to get it, but i didn't see that...
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  15. #15
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    x^2+4 helped, since it's located in the denominator. Then I thought in x^2-4, after subtraction I saw that I was doing worked out, besides x^2-4 it's the difference of perfect squares. Finally by seeing the x+2 in the denominator I could make it.

    I'm sorry if this explanation looks terrible, but it's pretty hard to explain what I did in a couple of seconds.
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