高等数学不定积分求解

Question

如题所述

Answer 1

令x = tanθ,dx = d(tanθ)
N = ∫ √(1 + x²) dx
= ∫ secθ d(tanθ)
= secθtanθ - ∫ tanθ d(secθ)
= secθtanθ - ∫ tan²θsecθ dθ
= secθtanθ - ∫ (sec²θ - 1)secθ dθ
= secθtanθ - N + ∫ secθ dθ
2N = secθtanθ + ln|secθ + tanθ| + C'
N = (x/2)√(1 + x²) + (1/2)ln|x + √(1 + x²)| + C
即∫ √(1 + x²) dx = (x/2)√(1 + x²) + (1/2)ln|x + √(1 + x²)| + C

Answer 2

Answer 3