Question

不定积分：∫1/(x^4+1)*dx=

Answer 1

1/(x^4+1)
=(-根号2＋1/2)/(x^2-根号2x+1)+(根号2/4+1/2)/(x^2+根号2+1)
∫1/(x^4+1)*dx
根号2/8积分（(2x+根号2）/(x^2+根号2+1)-(2x-根河2）/(x^2-根河2+1）dx+1/4积分[1/[(x+根河2/2)^2+(根河2/2)^2+1/[(x－根号2/2)^2+(根号2/2)^2]dx
=不会了！

Answer 2

∫1/(x^4+1)dx=
[1/(4*2^(1/2)]{ln[(x^2+2^(1/2)x+1)/(x^2-2^(1/2)x+1)]+2arctan[(2^(1/2)x)/(1-x^2)]}