Question

请问该怎么求解∫dx/((x²+1)(x²+x+1))呢？

我知道要用待定系数法，可总是解不出正确的系数(捂脸哭~

Answer 1

如图，剩下应该就好算了，望采纳

Answer 2

1/[(x²+1)(x²+x+1)] = (ax+b)/(x^2+1) + (cx+d)/(x^2+x+1)
= [(ax+b)(x^2+x+1)+(cx+d)(x^2+1)]/[(x²+1)(x²+x+1)]
a+c = 0, a+b+d = 0, a+b+c = 0, b+d = 1
解得 b = 0， d = 1， a = -1， c = 1
∫dx/[(x²+1)(x²+x+1)] = ∫[(x+1)/(x^2+x+1) - x/(x^2+1)]dx
= (1/2)∫[(2x+1+1)/(x^2+x+1) - (1/2)∫d(x^2+1)/(x^2+x+1)
= (1/2)ln(x^2+x+1) + (1/2)∫[d(x+1/2)/[(x+1/2)^2+3/4) - (1/2)ln(x^2+1)
= (1/2)ln[(x^2+x+1)/(x^2+1)] + (1/√3)arctan[(2x+1)/√3] + C

Answer 3

详细解答如下图片，关键是第一步的裂项变换

Answer 4

let

1/[(x^2+1)(x^2+x+1)] ≡ (Ax+B)/(x^2+1) + (Cx+D)/(x^2+x+1)
=>
1≡ (Ax+B)(x^2+x+1) + (Cx+D)(x^2+1)
coef. of x^3
A+C=0 (1)
coef. of x^2
A+B+D=0 (2)
coef. of x

A+B+C=0 (3)
coef. of constant
B+D=1 (4)
from (1) and (4)
A+B+D=0
A+1=0
A=-1
from (1)
A+C=0
-1+C=0
C=1
from (3)
A+B+C=0
-1+B+1=0
B=0
from (4)
B+D=1
0+D=1
D=1
ie
1/[(x^2+1)(x^2+x+1)] ≡ -x/(x^2+1) + (x+1)/(x^2+x+1)
∫dx/[(x^2+1)(x^2+x+1)]

=∫ [-x/(x^2+1) + (x+1)/(x^2+x+1) ] dx
=-(1/2)ln|x^2+1| +∫ (x+1)/(x^2+x+1) dx
=-(1/2)ln|x^2+1| +(1/2)∫ (2x+1)/(x^2+x+1) dx +(1/2)∫ dx/(x^2+x+1)
=-(1/2)ln|x^2+1| +(1/2)ln|x^2+x+1| +(1/2)∫ dx/(x^2+x+1)
=-(1/2)ln|x^2+1| +(1/2)ln|x^2+x+1| +(1/2)∫ dx/ [(x+1/2)^2 +(3/4)]
=-(1/2)ln|x^2+1| +(1/2)ln|x^2+x+1| +(2/3)∫ dx/ [1 +(4/3)(x+1/2)^2)]
=-(1/2)ln|x^2+1| +(1/2)ln|x^2+x+1| +(2/3)∫ dx/ {1 +[(2/√3)(x+1/2)]^2}
=-(1/2)ln|x^2+1| +(1/2)ln|x^2+x+1| +(√3/3)∫ d(2/√3)(x+1/2)/ {1 +[(2/√3)(x+1/2)]^2}
=-(1/2)ln|x^2+1| +(1/2)ln|x^2+x+1| +(√3/3)arctan[(2/√3)(x+1/2)] +C