赞 yangchunjiao 幼苗
共回答了13个问题采纳率：84.6% 举报
x/(1 - x³) = x/[(1 - x)(1 + x + x²)] = A/(1 - x) + (Bx + C)/(1 + x + x²)
x = A(1 + x + x²) + (Bx + C)(1 - x)
x = (A - B)x² + (A + B - C)x + (A + C)
A - B = 0 ==> B = A
A + B - C = 1
A + C = 0 ==> C = - A
A + B - C = 1
A + A + A = 1 ==> A = 1/3
B = 1/3,C = - 1/3
x/(1 - x³) = (x - 1)/[3(x² + x + 1)] - 1/[3(x - 1)]
∫ x/(1 - x³) dx
= (1/3)∫ (x - 1)/(x² + x + 1) dx - (1/3)∫ 1/(x - 1) dx
= (1/3)∫ [(1/2)(2x + 1 - 1) - 1]/(x² + x + 1) dx - (1/3)∫ 1/(x - 1) d(x - 1)
= (1/6)∫ (2x + 1)/(x² + x + 1) dx - (1/2)∫ 1/(x² + x + 1) dx - (1/3)Ln|x - 1|
= (1/6)∫ d(x² + x + 1)/(x² + x + 1) - (1/2)∫ 1/[(x + 1/2)² + 3/4] - (1/3)Ln|x - 1|
= (1/6)Ln|x² + x + 1| - (1/2)(2/√3)arctan[(x + 1/2) • 2/√3] - (1/3)Ln|x - 1| + C
= (1/6)Ln(x² + x + 1) - (1/3)Ln|x - 1| - (1/√3)arctan[(2x + 1)/√3] + C
x = A(1 + x + x²) + (Bx + C)(1 - x)
x = (A - B)x² + (A + B - C)x + (A + C)
A - B = 0 ==> B = A
A + B - C = 1
A + C = 0 ==> C = - A
A + B - C = 1
A + A + A = 1 ==> A = 1/3
B = 1/3,C = - 1/3
x/(1 - x³) = (x - 1)/[3(x² + x + 1)] - 1/[3(x - 1)]
∫ x/(1 - x³) dx
= (1/3)∫ (x - 1)/(x² + x + 1) dx - (1/3)∫ 1/(x - 1) dx
= (1/3)∫ [(1/2)(2x + 1 - 1) - 1]/(x² + x + 1) dx - (1/3)∫ 1/(x - 1) d(x - 1)
= (1/6)∫ (2x + 1)/(x² + x + 1) dx - (1/2)∫ 1/(x² + x + 1) dx - (1/3)Ln|x - 1|
= (1/6)∫ d(x² + x + 1)/(x² + x + 1) - (1/2)∫ 1/[(x + 1/2)² + 3/4] - (1/3)Ln|x - 1|
= (1/6)Ln|x² + x + 1| - (1/2)(2/√3)arctan[(x + 1/2) • 2/√3] - (1/3)Ln|x - 1| + C
= (1/6)Ln(x² + x + 1) - (1/3)Ln|x - 1| - (1/√3)arctan[(2x + 1)/√3] + C
1年前 追问
11
没有了，分母可以因式分解很好了 如果分子是x²的话还好凑微分
就是不定积分嘛，我都做得滚瓜烂熟了，加上软件验算也是这结果没错的 别把微积分当是个范畴，微积分只是积分学和微分学的总称而已
原来是你的格式不对了- - 原式是∫ x/(1 - x)³ dx嘛？怎么你会写成∫ x/(1 - x³) dx呢？3次方在括号内或外，过程有很大分别 令u = 1 - x，x = 1 - u，dx = - du ∫ x/(1 - x)³ dx = ∫ (1 - u)/u³ • (- du) = ∫ (u - 1)/u³ du = ∫ (1/u² - 1/u³) du = 1/(2u²) - 1/u + C = 1/[2(1 - x)²] - 1/(1 - x)] + C = (2x - 1)/[2(1 - x)²] + C
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你能帮帮他们吗