超难的不定积分,不信你试试!∫1／﹙1+x· tanx﹚²dx

设原函数 = A/(1 + xtanx) + C[(1 + xtanx)A' - A(tanx + xsec²x)]/(1 + xtanx)²令(1 + xtanx)A' - A(tanx + xsec²x) = 1若A = tanx,则A' = sec²x(1 + xtanx)(sec²x) - tanx(tanx + xsec...