求不定积分2x+4/x^2+2x+6 dx

1个回答

  • ∫(2x+4)/(x^2+2x+6) dx

    = ∫(2x+2)/(x^2+2x+6) dx + 2∫dx/(x^2+2x+6)

    =ln|x^2+2x+6| + 2∫dx/(x^2+2x+6)

    consider

    x^2+2x+6 = (x+1)^2 +5

    let

    x+1 = √5.tany

    dx =√5(secy)^2 dy

    ∫dx/(x^2+2x+6)

    =(√5/5)∫ dy

    =(√5/5)y

    =(√5/5)arctan[(x+1)/√5]

    ∫(2x+4)/(x^2+2x+6) dx

    =ln|x^2+2x+6| + 2∫dx/(x^2+2x+6)

    =ln|x^2+2x+6| + (2√5/5)arctan[(x+1)/√5] + C