∫x^2√1+x^2dx 积分要怎么算?

3个回答

  • 令x = tanθ,dx = sec²θ dθ

    ∫ x²√(1 + x²) dx

    = ∫ tan²θ * |secθ| * (sec²θ dθ)

    = ∫ tan²θsec³θ dθ

    = ∫ (sec²θ - 1)sec³θ dθ

    = ∫ sec⁵θ dθ - ∫ sec³θ dθ

    = (1/4)sec⁴θsinθ + (3/4)∫ sec³θ dθ - ∫ sec³θ dθ

    = (1/4)sec³θtanθ - (1/4)(1/2)[secθtanθ + ln|secθ + tanθ|] + C

    = (1/4)sec³θtanθ - (1/8)secθtanθ - (1/8)ln|secθ + tanθ| + C

    = (1/4)x(1 + x²)^(3/2) - (1/8)x√(1 + x²) - (1/8)ln|x + √(1 + x²)| + C

    = (x/8)(2x² + 1)√(1 + x²) - (1/8)ln|x + √(1 + x²)| + C

    公式,f(n) = ∫ (secx)^n dx

    f(n) = [(secx)^(n - 1) * sinx]/(n - 1) + (n - 2)/(n - 1) * f(n - 2)