partial fraction

Resolve [4x(x+4)] / (x^2 - 4) (x+2) into a partial fraction.

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  • 8 年前
    最愛解答

    Resolve [4x(x+4)]/[(x^2-4)(x+2)] into a partial fraction

    Sol

    [4x(x+4)]/[(x^2-4)(x+2)]=[4x(x+4)]/[(x-2)(x+2)^2]

    Set

    [4x(x+4)]/[(x^2-4)(x+2)]=a/(x-2)+b/(x+2)+c/(x+2)^2

    4x(x+4)=a(x+2)^2+b(x-2)(x+2)+c(x-2)

    when x=-2

    4*(-2)*2=c(-2-2)

    c=4

    4x^2+16x=a(x+2)^2+b(x-2)(x+2)+4x-8

    4x^2+12x+8=a(x+2)^2+b(x-2)(x+2)

    4x+4=a(x+2)+b(x-2)

    when x=-2

    -8+4=b(-2-2)

    b=1

    4x+4=a(x+2)+x-2

    3x+6=a(x+2)

    a=3

    So

    [4x(x+4)]/[(x^2-4)(x+2)]=3/(x-2)+1/(x+2)+4/(x+2)^2

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