Question

求有理函数不定积分法 2/x(x^2+1)dx

求详细步骤，谢谢

Answer 1

解：
∫2/[x(x²+1)] dx
=∫ [2/x-2/(x²+1)] dx
=2∫[1/x-1/(x²+1)] dx
=2(ln|x|-arctanx)+C

追问

请问第二步骤是怎么推导出来的？谢谢

追答

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

Answer 2

∫2/[x(x^2+1)] dx
=∫ [2/x-2x/(x^+1)] dx
=2∫[1/x-x/(x²+1)] dx
=2lnx-∫ 1/(x²+1)] dx^2
=2lnx-ln(1+x^2)+C