不定积分求解:∫(2x^2+2x+20)/[(x^2+2x+5)(x-1)]dx

1个回答

  • ∫{ (2x^2+2x+20)/[(x^2+2x+5)(x-1)] }dx

    = 2∫{ (x^2+2x+5)/[(x^2+2x+5)(x-1)] }dx - ∫ (2x-10)/[(x^2+2x+5)(x-1)] dx

    = 2∫ [1/(x-1) ]dx - ∫ (2x-10)/[(x^2+2x+5)(x-1)] dx

    let

    (2x-10)/[(x^2+2x+5)(x-1) ≡ A/(x-1) + (B1x+B2)/(x^2+2x+5)

    2x-10≡ A(x^2+2x+5) +(B1x+B2)(x-1)

    x=1

    8A=-8

    A=-1

    coef.of x^2

    A+B1=0

    B1=1

    coef.of constant

    5A-B2 =-10

    B2=5

    (2x-10)/[(x^2+2x+5)(x-1) ≡ -1/(x-1) + (x+5)/(x^2+2x+5)

    ∫{ (2x^2+2x+20)/[(x^2+2x+5)(x-1)] }dx

    = 2∫ [1/(x-1) ]dx - ∫ (2x-10)/[(x^2+2x+5)(x-1)] dx

    =2∫ [1/(x-1) ]dx + ∫[1/(x-1)]dx - ∫ (x+5)/(x^2+2x+5) dx

    =3ln|x-1| - ∫ (x+5)/(x^2+2x+5) dx

    =3ln|x-1| - (1/2)∫ (2x+2)/(x^2+2x+5) dx - 4∫ [1/(x^2+2x+5) ]dx

    =3ln|x-1| -(1/2)ln|(x^2+2x+5)| -4∫ [1/(x^2+2x+5) ]dx

    consider

    x^2+2x+5 = (x+1)^2 +4

    let

    x+1 = 2tany

    dx = 2(secy)^2 dy

    ∫ [1/(x^2+2x+5) ]dx

    =(1/2)∫ dy

    =y/2 + c'

    =(1/2)arctan[(x+1)/2] + C'

    ∫{ (2x^2+2x+20)/[(x^2+2x+5)(x-1)] }dx

    =3ln|x-1| -(1/2)ln|(x^2+2x+5)| -4∫ [1/(x^2+2x+5) ]dx

    ==3ln|x-1| -(1/2)ln|(x^2+2x+5)| -2arctan[(x+1)/2] + C