※ 引述《emily8877560 (Emily)》之銘言：
: 1-a
: 不定積分2t/(t^2+2t-3)^(1/2) dt
: 1-b
: 積分從零到拍 (2^1/2)/(3cosx+2) dx
: a題的關鍵是分母加2嗎?
: b題不知道從何做起
2t
∫---------------- dt
√(t^2 + 2t - 3)
2t
= ∫----------------- dt
√((t+1)^2 - 2^2)
2(2secθ - 1)
= ∫(---------------)(2)(secθ)(tanθ) dθ
2tanθ
(令 t + 1 = 2secθ , 則 dt = 2secθtanθ dθ)
= ∫(4)((secθ)^2) - (2)(secθ) dθ
= (4)(tanθ) - (2)(ln|secθ + tanθ|) + c_1
√(t^2 + 2t - 3) |t + 1 √(t^2 + 2t - 3)|
= (4)(----------------) - (2)(ln|----- + ----------------|) + c_1
2 | 2 2 |
= (2)(√(t^2 + 2t - 3)) - (2)(ln|t+1 + √(t^2 + 2t - 3)| - ln2) + c_1
= (2)(√(t^2 + 2t - 3)) - (2)(ln|t+1 + √(t^2 + 2t - 3)|) + c_1 + (2)(ln2)
= (2)(√(t^2 + 2t - 3)) - (2)(ln|t+1 + √(t^2 + 2t - 3)|) + c
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