an interesting integral

[@CodexMathematica]

a good part of mathematic

[@ahmedsabri-ng2xx]

an interesting integral

[@dan-florinchereches4892]

I would split the fractions.
First lets work on ploynomial on denominator
X^4=-1
X^2=+-i
X^2=i => x1=√2/2(1+i) x2=-√2/2(1+i)
X^2=-i => x3=√2/2(1-i)
X4=-√2/2(i-i).
Now x1*x3=1/2(1+1)=1 and x1+x3=√2
X2x4=1 and x2+x4=-√2
So x^4+1=(x^2+√2x+1)(x^2-√2x+1)
(Ax+B)/(X^2+√2x+1)+(Cx+D)/(X^2-√2x+1)=X^2/(x^4+1)
(Ax+B)(X^2-√2x+1)+(Cx+D)(X^2+√2x+1)=X^2
X=√2/2(1+i) => (D+C*√2/2(i+1))*(i+1+i+1)=+i
D(2+2i)+2√2C*i=i
D=0 C=1/2√2
Similarly then A=0 B=-1/2√2
=> X^2/(x^4+1)=√2/4(1/x^2-√2x+1)-1/(x^2+√2x+1))=√2/4(1/((x-√2/2)^2+(√2/2)^2)-1/((x+√2/2)^2+(√2/2)^2))
Now all that is left is using the identity
Arctan'(x/a)=a/(x^2+a^2)
I missed an x on nunerator so we need to first make the top similar to derivative of denominator first by afdind and subtracting a number which will resilt in logarithm antiderivative and then the arcatngent formula