Factor the denominator as follows
x^4+1=x^4+1+2x^2-2x^2=
(x^2+1)^2-2x^2=(x^2+sqrt2*x+1)*(x^2-sqrt2*x+1)
Hence we need to find constants A,B,C,D such as
(4x)/(x^4+1)=(A*x+B)/(x^2+sqrt2*x+1)+(C*x+D)/(x^2-sqrt2*x+1)
Giving four values to the variable x such as x={0,1,-1,2}
we form as a system with four equations and four variables A,B,C,D
solving this system we get the values of A,B,C,D
Finally we have that
(4 x)/(x^4+1) = -sqrt(2)/(x^2+sqrt(2) x+1)-sqrt(2)/(-x^2+sqrt(2) x-1)
Hence now we have to calculate the integral
int (4 x)/(x^4+1) dx= int -sqrt(2)/(x^2+sqrt(2) x+1)dx - int sqrt(2)/(-x^2+sqrt(2) x-1) dx
To calclate each of the integrals in the right side we need to bring them in the form of 1+[f(x)]^2 that is then integrable as arctangent.
After some calculations we have that
x² +sqrt2*x +1 =(1/2) { {sqrt2 [x + (1/sqrt2)]}² + 1}
and similarly
x² -sqrt2*x +1 =(1/2) { {sqrt2 [x - (1/sqrt2)]}² + 1}
Now remember that
int {d[f(x)]} / {[f(x)]²+ 1} dx= arctan[f(x)] + c
Thus in conclusion
int [(4x )/ (x⁴+1)] dx = - 2arctan[sqrt2*x + sqrt2(1/sqrt2)] + 2arctan[sqrt2x - sqrt2(1/sqrt2)] + c =
- 2arctan(sqrt2*x + 1) + 2arctan(sqrt2*x - 1) + c
