∫dx/[x(x^3+1)]
=∫dx/[x(x+1)(x^2-x+1)]
let
1/[x(x^3+1)]≡A/x +B/(x+1) + (Cx+D)/(x^2-x+1)
=>
1≡A(x+1)(x^2-x+1) +Bx(x^2-x+1)+ (Cx+D)x(x+1)
x=0, A=1
x=-1, B=-1/3
coef. of x^3
A+B+C=0
C=-2/3
coef. of x
B+D =0
D= 1/3
ie
1/[x(x^3+1)]≡1/x -(1/3)[1/(x+1)] -(1/3) (2x-1)/(x^2-x+1)
∫dx/[x(x^3+1)]
=∫{1/x -(1/3)[1/(x+1)] -(1/3) (2x-1)/(x^2-x+1)} dx
= lnx - (1/3)ln|x+1| - (1/3)ln|x^2-x+1| + Cf(x)=(1/3)ln|x^3/(1+x^3)|+C如何？=∫{1/x -(1/3)[1/(x+1)] -(1/3) (2x-1)/(x^2-x+1)} dx
= lnx - (1/3)ln|x+1| - (1/3)ln|x^2-x+1| + C
= ln|x^3/(1+x^3)|+C挺厲害的，不過
∫dx/x(x^3+1)就=(1/3)ln|x^3/(1+x^3)|+C啊

你沒有發(fā)現(xiàn)么。。。。。。