Question

1/【(1+x)(1+x∧2)】的不定积分

Answer 1

 1/[(1+x)(1+x^2)]

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

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

第一项：

第二项缺举：

-∫x/(1-x^4)dx=(1/2)∫1/(1-x^4)d(1-x^2)=(1/2)∫[1/(1+x^2)+1/(1-x^2)]d(1-x^2)

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

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

=(1/2)ln|(1-x^2)/(1+x^2)|+c

两项合在一起同类项合伏颂碧并后即答案了，樱槐太失败了这么繁杂，看来要好好复习一下。

Answer 2

∫扰姿 dx/[(1 + x)(1 + x²)]
= (1/2)∫ dx/(1 + x) + (1/2)∫ (1 - x)/(1 + x²) dx
= (1/2)∫ dx/(1 + x) + (1/2)∫ dx/(1 + x²) - (1/2)∫ x/(1 + x²桥孝) dx
= (1/2)∫ d(1 + x)/(1 + x) + (1/2)∫ dx/(1 + x²) - (1/4)∫ d(1 + x²)/(1 + x²)
= (1/2)ln(1 + x) + (1/2)arctanx - (1/4)ln(1 + x²) + C
= (1/2)arctanx + ln[√(1 + x)/(1 + x²)^¼] + C

分式分裂过程看下面：
_________________________________________________

令1/[(1 + x)(1 + x²)] = A/(1 + x) + (Bx + C)/(1 + x²)
1 = A(1 + x²) + (Bx + C)(1 + x)
1 = A + Ax² + Bx + Bx² + Cx + C
1 = (A + B)x² + (B + C)x + (A + C)
{ A + B = 0 → B = - A
{ B + C = 0 → C = - B = A
{ A + C = 1 → C = 1 - A
C = 1 - A → A = 1 - A → 2A = 1 → A = 1/2 = C，B = - 1/敏李稿2

追答

Answer 3

解：∫1/【(1+x)(1+x∧2)】dx
=∫1/2【袭枣1/敬笑(1+x)+1/(1+x^2)-x/(1+x^2)】dx
=1/2ln绝对值x+1/2arctanx-1/4ln（1+x^2)+C
事实上，学会怎么分裂分母就会积亮禅含分了吧？