求解这道不定积分题 急!1.∫ {(x^4 - 3x^2)/ [(x+5)^3] * (x-3)} dx = ?2 ∫

2个回答

  • 1,

    ∫ {(x^4 - 3x^2)/ [(x+5)^3] * (x-3)} dx =

    = ∫{x^5-3x^4 - 3x^3 + 9x^2)/ [(x+5)^3]}dx

    设g(x) = x^5 - 3x^4 - 3x^3 + 9x^2 = (x+5)^5 + a(x+5)^4 + b(x+5)^3 + c(x+5)^2 + e(x+5) + f,

    f=g(-5)=(-5)^5 - 3*5^4 - 3(-5)^3 + 9*5^2 = -8*5^4 + 3*5^3 + 9*5^2

    = -37*5^3 + 9*5^2 = (9-5*37)25= -4400.

    g'(x) = 5x^4 - 12x^3 - 9x^2 + 18x = 5(x+5)^4 + 4a(x+5)^3 + 3b(x+5)^2 + 2c(x+5) + e

    e = g'(-5) = 5*5^4 - 12(-5)^3 - 9*5^2 + 18(-5) = 37*5^3 - 9*5^2 - 90

    = -4400-90 = -4490

    g''(x) = 20x^3 - 36x^2 - 18x + 18 = 20(x+5)^3 + 12a(x+5)^2 + 6b(x+5) + 2c,

    c = g''(-5)/2 = 10(-5)^3 - 18*5^2 - 9(-5)+ 9 = -68*25 + 50 = -13550

    g'''(x) = 60x^2 - 72x - 18 = 60(x+5)^2 + 24a(x+5) + 6b,

    b = g'''(-5)/6 = 10*5^2 - 12(-5) - 3 = 250+60-3=307

    g^(4)(x) = 120x - 72 = 120(x+5) + 24a,

    a = g^(4)(-5)/24 = 5(-5)-3=-28

    ∫ {(x^4 - 3x^2)/ [(x+5)^3] * (x-3)} dx =

    = ∫{x^5-3x^4 - 3x^3 + 9x^2)/ [(x+5)^3]}dx

    = ∫{(x+5)^2-28(x+5)+307-13550/(x+5)-4490/(x+5)^2-4400/(x+5)^3}dx

    = (x+5)^3/3 -14(x+5)^2 + 307(x+5)-13550ln|x+5|+4490/(x+5)+2200/(x+5)^2 + C

    C = const.

    2,

    也按照题1的方法做,用待定系数法.

    要不,你自个练练?