求arctanx/(x2(1+x2))的不定积分?

2个回答

  • ∫arctanxdx/[x^2(1+x^2)]

    =∫arctanxdx/x^2 -∫arctanxdx/(1+x^2)

    =∫arctanxd(-1/x)-∫arctanxdarctanx

    =-(arctanx)/x +∫(1/x)darctanx-(arctanx)^2/2

    =-(arctanx)/x-(arctanx)^2/2+∫dx/[x(1+x^2)]

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

    原式=-(arctanx)/x-(arctanx)^2/2+lnx-(1/2)ln(1+x^2)+C