∫2/[x(x²+1)] dx
=∫ [2/x-2/(x²+1)] dx
=2∫[1/x-1/(x²+1)] dx
=2(ln|x|-arctanx)+C请问第二步骤是怎么推导出来的？谢谢可以令2/[x(x²+1)]=A/x-Bx/(x²+1)=(Ax²+A-Bx²)/[x(x²+1)]对比系数得：A-B=0，A=2，所以A=B=2所以∫2/[x(x²+1)] dx=∫ [2/x-2x/(x²+1)] dx=2∫1/x dx-∫ 2x/(x²+1)] dx=2∫1/x dx-∫ 1/(x²+1)] d(x²+1)=2ln|x|-ln(x²+1)+C 注意ln|x|不是lnx∫1/x dx=ln|x|+C