高等数学 不定积分？

做不定积分的题目时，一般需要对一些常见的函数的原函数、导函数熟练掌握，这样才能在解题时事半功倍。

let
1/[(x^2+1)(x^2+x)]≡ A/x +B/(x+1) +(Cx+D)/(x^2+1)
=>
1≡ A(x+1)(x^2+1) +Bx(x^2+1) +(Cx+D)x(x+1)
x=0, => A=1
x=-1, => B=-1/2
x=i
(Ci+D)i(i+1)=1
(Ci+D)(i-1)=1
(-C-D) +(-C+D)i =1
-C-D=1 (1)
-C+D=0 (2)
(1)+(2)
-2C =1
C=-1/2
from (2)
D=1/2
1/[(x^2+1)(x^2+x)]
≡ A/x +B/(x+1) +(Cx+D)/(x^2+1)
≡ 1/x -(1/2)[1/(x+1)] -(1/2)(x-1)/(x^2+1)
∫1/[(x^2+1)(x^2+x)]
=∫ [1/x -(1/2)[1/(x+1)] -(1/2)(x-1)/(x^2+1)] dx
=ln|x| -(1/2)ln|x+1| -(1/2)∫x/(x^2+1) dx +(1/2)∫dx/(x^2+1)
=ln|x| -(1/2)ln|x+1| -(1/4)ln|x^2+1| +(1/2)arctanx +C

用分式裂项法，

见下图

详细过程如图，希望能帮到你解决你心中的问题

希望写的很清楚