令t= tanx则 dx = d(arctant) = 1/(1+t²) dt原式= ∫ 1/(2tanx +3) dx= ∫ 1/(2t +3 ) * 1/(1+t²) dt1/[(2t +3 )(1+t²)] = A/(2t+3) + (Bt+C) / (1+t²) = [(A+2B)t² +(3B+2C)t +(A+3C)] / ...