1/2∫1/(2x-3)² d(2x-3)
=1/2×(-1/3) ×1/(2x-3)³+C
=-1/[6(2x-3)³] +C题打错了是∫dx/［1+（2x+3）］有没有平方的=1/2∫1/[1+(2x-3)²]d(2x-3)=1/2×arctan(2x-3)+c有平方题打错了是∫dx/［1+（2x+3）²］上面给你答案了呀=1/2∫1/[1+(2x+3)²]d(2x+3)=1/2×arctan(2x+3)+co