仅解释书中解法:
定义域: x ≥ 3 或 x ≤ -3.
当 x ≥ 3 时, 令 x = 3secu,则 0 ≤ u ≤ π/2,得
I = ∫3tanu 3secutanudu / (3secu) = 3∫(tanu)^2du
= 3∫[(secu)^2-1]du = 3tanu - 3u + C
= √(x^2-9) - 3arccos(3/x) + C
当 x ≤ -3 时, 令 x = 3secu,则 π/2 ≤ u ≤ π,得
I = ∫(-3tanu) 3secutanudu / (3secu) = -3∫(tanu)^2du
= -3∫[(secu)^2-1]du = -3tanu + 3u + C
= √(x^2-9) + 3arccos(3/x) + C
因 x ≤ -3, arccos(3/x) = π - arccos[3/(-x)]
例如 x = -6, arccos(3/x) = arccos(-1/2) = 2π/3,
arccos[3/(-x)] = arccos(1/2) = π/3,
则 arccos(3/x) = π - arccos[3/(-x)] = 2π/3。
I = √(x^2-9) - 3arccos[3/(-x)] + 3π + C本回答被提问者采纳