Re: [積分] 一題不定積分
x^3 + 4x^2 - 4x - 1
∫--------------------- dx
(x^2 + 1)^2
(x^3 - 1) + (4x^2 - 4x) (x - 1)(x^2 + x + 1) + 4x(x - 1)
= ∫----------------------- dx = ∫----------------------------------- dx
(x^2 + 1)^2 (x^2 + 1)^2
(x - 1)(x^2 + 5x + 1) (x - 1)(x^2 + 1 ) 5x(x - 1)
= ∫----------------------- dx = ∫------------------ dx + ∫----------- dx
(x^2 + 1)^2 (x^2 + 1)^2 (x^2 + 1)^2
x - 1 5x(x - 1)
= ∫----------- dx + ∫----------- dx
x^2 + 1 (x^2 + 1)^2
Let x = tanθ dx = (secθ)^2 dθ
tanθ - 1 5(tanθ)^2 - 5tanθ
= ∫------------- (secθ)^2 dθ + ∫--------------------- (secθ)^2 dθ
(secθ)^2 (secθ)^4
= ∫ (tanθ - 1) dθ + ∫ [ 5(cosθ)^2 - 5sinθcosθ] dθ
5
= ln |secθ| - θ + --- [ 2θ + cos(2θ) - sin(2θ) ] + C
4
1 3 -1 5(1 - x)
= --- ln |x^2 + 1| + --- tan x + ----------- + C
2 2 2(x^2 + 1)
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希望有幫到你的忙 :D
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09/23 19:02, , 1F
09/23 19:02, 1F
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