窦靖童
幼苗
共回答了26个问题采纳率:84.6% 举报
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
1年前
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