求定积分，拆项。

Answer 1

这种拆项要用待定系数法，前一项分子变为a，后一项也就是分母是二次项那个，分子写为bx+c，然后通分，确定abc的值

Answer 2

∫ dx/[x(1+x^2)]
let
1/[x(1+x^2)]≡ A/x + (B1x+B2)/(1+x^2)
=>
1≡ A(1+x^2)+ x(B1x+B2)

coef. of x, =>B2 =0
x=0, => A=1

coef. of x^2
A+B1=0
B1 =-1
ie
1/[x(1+x^2)]≡ 1/x - x/(1+x^2)

∫ dx/[x(1+x^2)]
=∫ [1/x - x/(1+x^2)] dx
=ln|x| - (1/2)ln(1+x^2) + C

Answer 3

∫ dx/[x(1+x^2)]
let
1/[x(1+x^2)]≡ A/x + (B1x+B2)/(1+x^2)
=>
1≡ A(1+x^2)+ x(B1x+B2)

coef. of x, =>B2 =0
x=0, => A=1

coef. of x^2
A+B1=0
B1 =-1
ie
1/[x(1+x^2)]≡ 1/x - x/(1+x^2)

∫ dx/[x(1+x^2)]
=∫ [1/x - x/(1+x^2)] dx
=ln|x| - (1/2)ln(1+x^2) + C

Answer 4

解：∵1/(x(1+x^2))=1/x-x/((1+x^2)，∴不定积分=ln丨x丨-（1/2）ln(1+x^2)+c。供参考啊。