Perform the decomposition into partial fractions
(x-1)/(x(x^4+1))=(Ax^3+Bx^2+C)/(x^4+1)+(D)/(x)
=((Ax^3+Bx^2+C)x+D(x^4+1))/(x(x^4+1))
The denominators are the same, compare the numerators
x-1=Ax^4+Bx^3+Cx+Dx^4+D
Therefore, comparing the LHS and the RHS
A+D=0
C=1
D=-1
B=0
A=-D=1
So,
(x-1)/(x(x^4+1))=(x^3+1)/(x^4+1)-1/(x)
int((x-1)dx)/(x(x^4+1))=int((x^3+1)dx)/(x^4+1)-int(1dx)/(x)
=int(x^3dx)/(x^4+1)+int(dx)/(x^4+1)-int(1dx)/(x)
Let u=x^4+1, =>, du=4x^3dx
Therefore,
int(x^3dx)/(x^4+1)=1/4int(du)/(u)=1/4lnu=1/4ln(x^4+1)
int(dx)/(x)=ln(|x|)
1/(x^4+1)=1/((x^2-sqrt2x+1)(x^2+sqrt2x+1))
=(Ax+B)/(x^2-sqrt2x+1)+(Cx+D)/(x^2+sqrt2x+1)
=((Ax+B)(x^2+sqrt2x+1)+(Cx+D)(x^2-sqrt2x+1))/((x^2-sqrt2x+1)(x^2+sqrt2x+1))
The denominators are the same, we compare the numerators
1=(Ax+B)(x^2+sqrt2x+1)+(Cx+D)(x^2-sqrt2x+1)
Let x=0, =>, 1=B+D
Coefficients of x^3, =>, A+C=0
Coefficients of x, =>, sqrt2B+A+C-sqrt2D=0
=>, B=D, =>,B=D=1/2
Coefficients of x^2, =>, Asqrt2+B-sqrt2C+D=0
=>, A=-1/(2sqrt2)
=>, C=1/(2sqrt2)
So,
1/(x^4+1)=(-1/(2sqrt2)x+1/2)/(x^2-sqrt2x+1)+(1/(2sqrt2)x+1/2)/(x^2+sqrt2x+1)
=(sqrt2-x)/(2sqrt2(x^2-sqrt2x+1))+(sqrt2+x)/(2sqrt2(x^2+sqrt2x+1))
int(dx)/(x^4+1)=int((sqrt2-x)dx)/(2sqrt2(x^2-sqrt2x+1))+int((sqrt2+x)dx)/(2sqrt2(x^2+sqrt2x+1))
int((sqrt2-x)dx)/(2sqrt2(x^2-sqrt2x+1))=1/(4sqrt2)(-ln|2x^2-2sqrt2x+2|)+2arctan(2x-1)
int((sqrt2+x)dx)/(2sqrt2(x^2+sqrt2x+1))=1/(4sqrt2)(ln|2x^2+2sqrt2x+2|)+2arctan(2x+1)
