#x^4+1# is not able to be factorized in rationals.
However, like #x^4+4=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)#, you can factorize #x^4+1# in reals:
#x^4+1=(x^2+1)^2-(sqrt(2)x)^2=(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1)#
Note that #x^2+-sqrt(2)x+1=(x+-sqrt(2)/2)^2+1/2# ・・(A) cannot be factorized in reals anymore.
Thus, #1/(x^4+1)# can be rewritten in partial fractions form:
#1/(x^4+1)=(ax+b)/(x^2+sqrt(2)x+1)+(cx+d)/(x^2-sqrt(2)x+1)#・・(B)
(#a,b,c,dinRR#)
What we want to do is to find #a,b,c,d# so that (B) is an identity.
Multyply both sides by #x^4+1# and you find
#1=(ax+b)(x^2-sqrt(2)x+1)+(cx+d)(x^2+sqrt(2)x+1)#
#1=(a+c)x^3+(-sqrt(2)a+b+sqrt(2)c+d)x^2+(a-sqrt(2)b+c+sqrt(2)d)x+(b+d)#
Make a comparison of coefficients and you will see:
#a+c=0# (i)
#-sqrt(2)a+b+sqrt(2)c+d=0# (ii)
#a-sqrt(2)b+c+sqrt(2)d=0# (iii)
#b+d=1# (iv)
By (i) and (iv) you can easily see:
#c=-a#, #d=1-b#
Substitute these in (ii) and (iii):
#-2sqrt(2)a+1=0#, #-2sqrt(2)b+sqrt(2)=0#
#a=sqrt(2)/4, b=1/2, c=-sqrt(2)/4, d=1/2#
We have completed to simplify (B) and got to the start line.
#color(blue)(1/(x^4+1)=(sqrt(2)/4x+1/2)/(x^2+sqrt(2)x+1)+(-sqrt(2)/4x+1/2)/(x^2-sqrt(2)x+1))#
Now let's go to the next stage: integration.
Since #d/(dx)(x^2+sqrt(2)x+1)=2x+sqrt(2)# and #sqrt(2)/4x+1/2=sqrt(2)/8(2x+sqrt(2))+1/4#,
#int((sqrt(2)/4x+1/2)/(x^2+sqrt(2)x+1))dx#
#=sqrt(2)/8int(2x+sqrt(2))/(x^2+sqrt(2)x+1)dx+1/4int1/(x^2+sqrt(2)x+1)dx#
#=sqrt(2)/8ln(x^2+sqrt(2)x+1)+sqrt(2)/4arctan(sqrt(2)x+1)+C#.
Note that #color(green)(int1/((x+p)^2+q^2)dx=arctan((x+p)/q)/q)# for #q>0#. Here, #p=q=sqrt(2)/2# (See (A)).
In similar ways (this will be a good excercise to show this^^),
#int((-sqrt(2)/4x+1/2)/(x^2-sqrt(2)x+1))dx#
#=-sqrt(2)/8ln(x^2-sqrt(2)x+1)-sqrt(2)/4arctan(-sqrt(2)x+1)+C#.
Sum up the two integration and you will finally acheived the goal:
#color(red)(int1/(x^4+1)dx=sqrt(2)/8ln((x^2+sqrt(2)x+1)/(x^2-sqrt(2)x+1))+sqrt(2)/4(arctan(sqrt(2)x+1)-arctan(-sqrt(2)x+1))+C)#.
