求不定积分 1/(X^2+1)*arctan[(1+x)/(1-x)]dx

分部积分,不要嫌麻烦,很简单的,你试试
原始=积arctan[(1+x)/(1-x)]darctanx=
arctan[(1+x)/(1-x)]*arctanx-积arctanxdarctan[(1+x)/(1-x)]=arctan[(1+x)/(1-x)]*tanx-积arctanxdarctanx=arctan[(1+x)/(1-x)]*tanx-1/2（arctanx）^2+C