求 (x^2 +1)⼀(x^4 +1) 的不定积分~

2024-11-11 18:24:25
推荐回答(1个)
回答1:

S(x^2 +1)/(x^4 +1)dx
=S(1+1/x^2)/(x^2+1/x^2)dx
=S1/[(x-1/x)^2+2] d(x-1/册竖x)
t=x-1/x
=S1/(t^2+2) dt
=根2/猜姿橡2*arctan(t/根2)+c
=根穗旁2/2*arctan((x-1/x)/根2)+c