∫x^2√1+x^2dx 积分要怎么算?

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)

let
x= tana
dx = (seca)^2 da
∫x^2√(1+x^2)dx
= ∫ (tana)^3 (seca)^2 da
= ∫ ((seca)^2-1)(seca)d (seca)
= (seca)^4/4 - (seca)^2/2 +C
= (x^2+1)^4/4 - (x^2+1)^2/2 + C

令x = tanθ，dx = sec²θ dθ。
∫ x²√(1 + x²) dx
= ∫ tan²θ * |secθ| * (sec²θ dθ)
= ∫ tan²θsec³θ dθ
= ∫ (sec²θ - 1)sec³θ dθ
= ∫ sec⁵θ dθ ...