第3小题,第三题第1小题

1个回答

  • ∫[(x+3)/(x^2+2x+10)]dx

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

    =(1/2)×∫[d(x^2+2x+10)/(x^2+2x+10)]+2∫[1/(x^2+2x+10)]dx

    =(1/2)ln(x^2+2x+10)+2∫[1/(x+1)^2+3^2]dx

    =(1/2)ln(x^2+2x+10)+(2/3)arctan[(x+1)/3]+C

    ∫[cosx/√(2+cos2x)]dx

    =∫[cosx/√(2+1-2sin^2 x)]dx

    =∫[1/√(3-2sin^2 x)]d(sinx)

    =(1/√2)×∫{1/√[(√3)^2-(√2sinx)^2]}d(√2sinx)

    =(1/√2)×(1/√3)×arcsin[√2sinx/√3]

    =(√6/6)arcsin[(√6/3)sinx]+C