3/（x^3+1) 和x^2+1/x^4+1 求积分。。。

来源：百度知道 编辑：UC知道 时间：2024/05/11 02:04:22
麻烦路过的大侠。。。写下具体过程。。。
帮帮忙啦。。。我想到头都破了。。。。

3/(x^3 + 1)

= 3/[(x+1)(x^2 - x + 1)]

= 3[a/(x+1) + (bx+c)/(x^2 - x + 1)]

a(x^2 - x + 1) + (bx+c)(x+1)

= ax^2 - ax + a + bx^2 + cx + bx + c

= (a+b)x^2 + (b+c-a)x + (a+c)

= 1

a + b = 0,
b + c - a = 0,
a + c = 1,

a = 1/3, b = -1/3, c = 2/3.

3/(x^3 + 1)

= 3[a/(x+1) + (bx+c)/(x^2 - x + 1)]

= 1/(x+1) + (2-x)/(x^2 - x + 1)

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

= 1/(x+1) + (-1/2)(2x-1)/(x^2 - x + 1) + (3/2)/[(x - 1/2)^2 + 3/4]

= 1/(x+1) + (-1/2)(2x-1)/(x^2 - x + 1) + 3^(1/2)[2/3^(1/2)]/{[2/3^(1/2)(x - 1/2)]^2 + 1}

积分为，
ln|x+1| - 1/2ln(x^2 - x + 1) + 3^(1/2)arctan[(2x-1)/3^(1/2)] + C

(x^2 + 1)dx/(x^4 + 1)

= (1 + 1/x^2)dx/(x^2 + 1/x^2)

= d(x - 1/x)/[(x - 1/x)^2 + 2]

= 2^(-1/2)d[(x - 1/x)/2^(1/2)]/{[(x - 1/x)/2^(1/2)]^2 + 1}

积分