We need
#intdx/(x^2+1)=arctan(x)#
As the degree of the numerator is greater than the degree of the denominator, perform a long division first.
#(x^5+3x^2+1)/(x^4-1)=x+(3x^2+x+1)/(x^4-1)#
Perform a decomposition into partial fractions
#(3x^2+x+1)/(x^4-1)=(3x^2+x+1)/((x^2+1)(x+1)(x-1))#
#=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)#
#=((Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1))/((x^2+1)(x+1)(x-1))#
The denominators are the same , compare the numerators
#3x^2+x+1=(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)#
Let #x=1#, #=>#, #5=4D#, #=>#, #D=5/4#
Let #x=-1#, #=>#, #3=-4C#, #=>#, #C=-3/4#
Let #x=0#, #=>#, #1=-B-C+D#, #=>#, #B=5/4+3/4-1=1#
Coefficients of #x^3#
#0=A+C+D#, #=>#, #A=-C-D=3/4-5/4=-1/2#
Therefore,
#(x^5+3x^2+1)/(x^4-1)=x+(-1/2x+1)/(x^2+1)+(-3/4)/(x+1)+(5/4)/(x-1)#
So,
#int((x^5+3x^2+1)dx)/(x^4-1)=intxdx+int((-1/2x+1)dx)/(x^2+1)+int(-3/4dx)/(x+1)+int(5/4dx)/(x-1)#
#intxdx=x^2/2#
#int(-3/4dx)/(x+1)=-3/4ln(|x+1|)#
#int(5/4dx)/(x-1)=5/4ln(|x-1|)#
#int((-1/2x+1)dx)/(x^2+1)=-1/4int(2xdx)/(x^2+1)+intdx/(x^2+1)=-1/4ln(x^2+1)+arctanx#
Finally,
#int((x^5+3x^2+1)dx)/(x^4-1)=x^2/2-3/4ln(|x+1|)+5/4ln(|x-1|)-1/4ln(x^2+1)+arctanx+C#
