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Evaluate Integral x^2+4 / x^4+x^2+16 dx

  • Math -

    gotta get creative with this one. Note that

    x^4+x^2+16 = x^4-8x^2+16 - 9x^2 = (x^2-4)^2 - (3x)^2

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

    so, if you let u = 3x/(x^2-4)
    du = -3(x^2+4)/(x^2-4)^2 and you have

    integral -1/3 du/(1+u^2)

    take it from there!

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