A.maths question~

我假設閣下所指是:
∫[2x/(3x + 1)-1/2]dx = ∫2x √(3x + 1) dx
則需用到代入法 (Method of substitution) 解 (此方式在今日的 A.M. 已 out C)
代 u = √(3x + 1), 則 du = 1/[2√(3x + 1)] d(3x + 1) = 3dx/(2u)
所以: dx = [2√(3x + 1)]du/3 = 2udu/3
而 x = (u2 - 1)/3
則:
∫2x √(3x + 1) dx = ∫[2u (u2 - 1)/3] (2u du/3)
= (4/9) ∫u2 (u2 - 1) du
= (4/9) ∫(u4 - u2) du
= (4/9) (u5/5 - u3/3) + C
= (4u/135) (3u4 - 5u2) + C
= [4√(3x + 1)/135] [3(3x + 1)2 - 5(3x + 1)]
= [4√(3x + 1)/135] [3(9x2 + 6x + 1) - 5(3x + 1)]
= 4(27x2 + 3x - 2)[√(3x + 1)]/135
= 4(9x - 2)(3x + 1)[√(3x + 1)]/135
= 4(9x - 2)(3x + 1)3/2/135

感謝六呎將軍的另外method~但OUT　ｃ方法５行喇＝　。　＝

不用method of substitution也可解
只需考慮
2x(3x + 1)^0.5
= 2/3 [3x(3x + 1)^0.5]
= 2/3 [(3x + 1 - 1)(3x + 1)^0.5]
= 2/3 (3x + 1)^1.5 - 2/3 (3x + 1)^0.5