求不定积分,

1个回答

  • ∴∫√(x^2-2)dx

    =(1/2)∫[2√(x^2-2)]dx

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

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

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

    =(1/2)∫d[x√(x^2-2)]-∫[{x′+[√(x^2-2)]′}/[x+√(x^2-2)]]dx

    =(1/2)x√(x^2-2)-∫d{ln[x+√(x^2-2)]}

    =(1/2)x√(x^2-2)-ln[x+√(x^2-2)]+C.