∫3x^3/(x^2 + 1)^3dx 麻煩看看我有無計錯

我分別用了三角代入法，以及普通的代入法，去計算 "∫ 3x^3/(x^2 + 1)^3 dx"。兩個答案不同，為什麼？麻煩看看我有無計錯：

三角代入法：

∫ 3x^3/(x^2 + 1)^3 dx

= ∫ 3x^3/(1^2 + x^2)^3 dx

Let x = tanθ. (trigonometric substitution)

dx/dθ = sec^2θ

dx = sec^2θ dθ

= ∫ 3 * tan^3θ/(1 + tan^2θ)^3 * sec^2θ dθ

= ∫ 3 * tan^3θ/sec^6θ * sec^2θ dθ

= ∫ 3 * tan^3θ/sec^4θ dθ

= ∫ 3 * sin^3θcosθ dθ

Let u = sinθ.

du/dθ = cosθ

dθ = du/cosθ

= ∫ 3 * u^3 du

= 3/4 * u^4 + C

= 3/4 * sin^4θ + C

= 3/4 * (x/√(1 + x^2))^4 + C

= 3/4 * x^4/(1 + x^2)^2 + C

普通的代入法：

∫ 3x^3/(x^2 + 1)^3 dx

Let u = x^2 + 1.

du/dx = 2x

dx = du/2x

= ∫ 3x^3/u^3 du/2x

= ∫ 3/2 * x^2/u^3 du

= ∫ 3/2 * (u - 1)/u^3 du

(u - 1)/u^3 = A/u + B/u^2 + C/u^3 (partial fraction)

u - 1 = Au^2 + Bu + C

A = 0

B = 1

C = -1

= 3/2 * ∫ 1/u^2 - 1/u^3 du

= 3/2 * (-1 * 1/u - 1/(-2) * 1/u^2) + C

= 3/2 * (1/2 * 1/u^2 - 1/u) + C

= 3/2 * (1/(2u^2) - 2u/(2u^2)) + C

= 3/2 * (1 - 2u)/(2u^2) + C

= 3/4 * (1 - 2u)/u^2 + C

= 3/4 * (1 - 2(x^2 + 1))/(x^2 + 1)^2 + C

= 3/4 * (-2x^2 - 1)/(x^2 + 1)^2 + C

= -3/4 * (2x^2 + 1)/(x^2 + 1)^2 + C