Here.
#I=int1/(1+x^4)dx#
#=1/2[int((x^2+1)-(x^2-1))/(x^4+1)dx]#
#=1/2int(x^2+1)/(x^4+1)dx-1/2int(x^2-1)/(x^4+1)dx#
#=1/2int(1+1/x^2)/(x^2+1/x^2)dx-1/2int(1-1/x^2)/(x^2+1/x^2)dx#
#=1/2I_1-1/2I_2...to(A)#
Now , #I_1=int(1+1/x^2)/((x^2+1/x^2)dx#
#I_1=int((1+1/x^2))/((x-1/x)^2+2)dx#
Subst. #x-1/x=u=>(1+1/x^2)dx=du#
#=>I_1=int1/(u^2+(sqrt2)^2)du##=1/sqrt2arc tan(u/sqrt2)+c_1#
#=>I_1=1/sqrt2arc tan((x-1/x)/sqrt2)+c_1 ,to[becauseu=x-1/x]#
#=>I_1=1/sqrt2arc tan((x^2-1)/(sqrt2x))+c_1...to(B)#
Again ,we have
#I_2=int(1-1/x^2)/(x^2+1/x^2)dx#
#I_2=int((1-1/x^2))/((x+1/x)^2-2)dx#
Subst. #x+1/x=v=>(1-1/x^2)dx=dv#
#=>I_2=int1/(v^2-(sqrt2)^2)dv=1/(2sqrt2)ln|(v-sqrt2)/(v+sqrt2)
|+c_2#
#=>I_2=1/(2sqrt2)ln|(x+1/x-sqrt2)/(x+1/x+sqrt2)|+c_2 ,to[becausev=x+1/x]#
#=>I_2=1/(2sqrt2)ln|(x^2-sqrt2x+1)/(x^2+sqrt2x+1)|+c_2...to(C)#
From #(A), (B) ,and (C)# we get
#I#=#1/2*1/sqrt2arc tan((x^2-1)/(sqrt2x))-1/2*1/(2sqrt2)ln|
(x^2-sqrt2x+1)/(x^2+sqrt2x+1)|+C#
#I#=#1/(2sqrt2)arc tan((x^2-1)/(sqrt2x))-1/(4sqrt2)ln|
(x^2-sqrt2x+1)/(x^2+sqrt2x+1)|+C#
Where #C=1/2(c_1-c_2)#
