求不定积分 ∫dx⼀(1+ x눀)눀

2024-11-07 04:36:25
推荐回答(1个)
回答(1):

令:x=tant
则:dx=sec²t
∫dx/(1+x²)²
=∫sec²tdt/(1+tan²)²
=∫1/sec²t dt
=∫cos²t dt
=∫1/2+(1/2)*cos2t dt
=t/2+(1/4)*sin2t+C
再把t换回去,t=arctanx
结果为arctanx/2+(1/4)*sin2arctanx+C