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    Hi there.
    I'm stuck in this qustion

    \frac{ax+b}{x^2+2x+2}+\frac{cx+d  }{x^2-2x+2}=\frac{1}{x^4+4}

    I'm having trouble working out the values of a,b,c,d.

    I did:
    (ax+b)(x^2-2x+2)+(cx+d)(x^2+2x+2)≡1
    and got when x=0
    (i) 2b+2d≡1
    (ii)a=-c;
    (iii)-2a+b+2c+d≡0;
    (iv)2a-2b+2c+2d≡0

    so c=-1/8; a=1/8
    d=b=1/4

    This doesn't seem right because i need to integrate the two fractions and it doesn't work.

    Thanks.
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    (Original post by Goldenratio)
    Hi there.
    I'm stuck in this qustion

    \frac{ax+b}{x^2+2x+2}+\frac{cx+d  }{x^2-2x+2}=\frac{1}{x^4+4}

    I'm having trouble working out the values of a,b,c,d.

    I did:
    (ax+b)(x^2-2x+2)+(cx+d)(x^2+2x+2)≡1
    and got when x=0
    (i) 2b+2d≡1
    (ii)a=-c;
    (iii)-2a+b+2c+d≡0;
    (iv)2a-2b+2c+2d≡0

    so c=-1/8; a=1/8
    d=b=1/4

    This doesn't seem right because i need to integrate the two fractions and it doesn't work.

    Thanks.
    You can integrate the LHS whatever a,b,c,d are.

    Make a sub of u = x+1 in the left fraction and v = x-1 in the right and it might become clearer.
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    Thanks!
    Sorry I'm not seeing it.
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    (Original post by Goldenratio)
    Thanks!
    Sorry I'm not seeing it.
    (ax+b)/(x^2+2x+2) becomes

    au/(u^2+1) + (b-a)/(u^2+1)

    which are standard integrals.
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    How did you split the fraction? Obviously there arent any factors so...
    It's a STEP ii question, but i'm just not getting there. :p:
    Thanks
    • Thread Starter
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    Why couldn't I integrate the 1/(x^4+4) by substituting x^2=2tan^2p?
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    (Original post by Goldenratio)
    How did you split the fraction? Obviously there arent any factors so...
    What do you mean? They split the fraction for you :confused:

    I just made the u = x+1 sub I'd previously suggested.
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    Okay thanks, got you!
 
 
 
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Updated: August 28, 2005

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