求函数f(x)=(4x^2+1)^0.5的不定积分

Question

求函数f(x)=(4x^2+1)^0.5的不定积分

Answer 1

令2x = tanθ,2 dx = sec²θ dθ,4x² = tan²θ
∫ √(4x² + 1) dx
= ∫ √(tan²θ + 1) · (1/2)sec²θ dθ
= (1/2)∫ sec³θ dθ
= (1/2)∫ secθ dtanθ
= (1/2)secθtanθ - (1/2)∫ tanθ dsecθ
= (1/2)secθtanθ - (1/2)∫ secθtan²θ dθ
= (1/2)secθtanθ - (1/2)∫ secθ(sec²θ - 1) dθ
= (1/2)secθtanθ - (1/2)∫ sec³θ dθ + (1/2)∫ secθ dθ
∫ sec³θ dθ = (1/2)secθtanθ + (1/2)ln|secθ + tanθ| + C
(1/2)∫ sec³θ dθ = (1/4)(2x)√(4x² + 1) + (1/4)ln|2x + √(4x² + 1)| + C
====> ∫ √(4x² + 1) dx = (x/2)√(4x² + 1) + (1/4)ln|2x + √(4x² + 1)| + C