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    x^2 + x / (x^2 - 1)(x^2 + 1)


    x^2+x = A(x^2 + 1) + B(x^2 - 1)

    Let x = 1

    A=1

    How do I find B?

    Thanks
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    (Original post by khanpatel321)
    x^2 + x / (x^2 - 1)(x^2 + 1)


    x^2+x = A(x^2 + 1) + B(x^2 - 1)

    Let x = 1

    A=1

    How do I find B?

    Thanks
    x2 - 1 factorizes further
    x2+1 is irreducible
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    (Original post by khanpatel321)
    x^2 + x / (x^2 - 1)(x^2 + 1)


    x^2+x = A(x^2 + 1) + B(x^2 - 1)

    Let x = 1

    A=1

    How do I find B?

    Thanks
    You want \frac{Ax+B}{x^2+1}+\frac{C}{x+1}  +\frac{D}{x-1}
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    (Original post by TeeEm)
    x2 - 1 factorizes further
    x2+1 is irreducible
    (Original post by brianeverit)
    You want \frac{Ax+B}{x^2+1}+\frac{C}{x+1}  +\frac{D}{x-1}
    I thought you only add a polynomial when the fraction is an improper fraction?

    Why isn't it

    A/(x^2 + 1) + B/(x-1) + C/(x+1)
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    (Original post by khanpatel321)
    I thought you only add a polynomial when the fraction is an improper fraction?

    Why isn't it

    A/(x^2 + 1) + B/(x-1) + C/(x+1)
    brianeverit has given you the correct form ...

    the reason for what you are asking you may not understand at present but here it comes...
    when you divide by an irreducible quadratic the most general form for the remainder is a linear expression
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    (Original post by TeeEm)
    brianeverit has given you the correct form ...

    the reason for what you are asking you may not understand at present but here it comes...
    when you divide by an irreducible quadratic the most general form for the remainder is a linear expression
    (Original post by khanpatel321)
    I thought you only add a polynomial when the fraction is an improper fraction?

    Why isn't it

    A/(x^2 + 1) + B/(x-1) + C/(x+1)
    (Original post by brianeverit)
    You want \frac{Ax+B}{x^2+1}+\frac{C}{x+1}  +\frac{D}{x-1}

    Ok I understand now. Just one last question... when differentiating sin(2x) the answer is 2cos(2x) but when you differentiate 2sin(-2x) the answer is -2cos(2x) why isn't it -2cos(-2x) ?
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    (Original post by khanpatel321)
    Ok I understand now. Just one last question... when differentiating sin(2x) the answer is 2cos(2x) but when you differentiate 2sin(-2x) the answer is -2cos(2x) why isn't it -2cos(-2x) ?
    cos is even

    cos(2x) is the same as cos(-2x)
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    (Original post by TeeEm)
    cos is even

    cos(2x) is the same as cos(-2x)
    e^(-x) sin(-2x)

    d/dx = -e^(-x)sin(-2x) - 2e(^-x)cos(-2x)

    is this correct?
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    (Original post by khanpatel321)
    e^(-x) sin(-2x)

    d/dx = -e^(-x)sin(-2x) - 2e(^-x)cos(-2x)

    is this correct?
    It is but I would have written

    d/dx = e^(-x)sin(2x) - 2e(^-x)cos(2x)
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    (Original post by khanpatel321)
    Ok I understand now. Just one last question... when differentiating sin(2x) the answer is 2cos(2x) but when you differentiate 2sin(-2x) the answer is -2cos(2x) why isn't it -2cos(-2x) ?
    \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigg[2 \sin (-2x)\bigg] = -4 \cos (2x)

    Since cosine is an even function. That is f(x) = \cos x has the property that f(x) = f(-x).
 
 
 
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