Before we rewrite #x^4/(x^4-1)# into partial fractions, first make sure that the degree of the denominator is strictly greater than the degree of the numerator.
So, we can use #x^4/(x^4-1)=(x^4-1+1)/(x^4-1)=1+1/(x^4-1)#
Now, we rewrite #1/(x^4-1)# using partial fractions. First, factorize #x^4-1=(x^2+1)(x+1)(x-1)#.
Thus, we know that #1/(x^4-1)# can be rewritten in the form #(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)#.
Multiply both sides by #x^4-1# to get #1=(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)#.
Set #x=1# to get #D=1/4# and #x=-1# to get #C=-1/4#
Multiply this out to get #(A+C+D)x^3+(B-C+D)x^2-(A-C-D)x-B-C+D=1#.
Substitute #D=1/4# and #C=-1/4# to finally get #Ax^3+(B+1/2)x^2-Ax-B+1/2=1#.
Equate coefficients: #{(A=0),(B+1/2=0),(A=0),(-B+1/2=1):}#.
From the above, we get #{(A=0),(B=-1/2),(C=-1/4),(D=1/4):}#.
Thus, we know that #1/(x^4-1)=-1/(2x^2+2)-1/(4x+4)+1/(4x-4)#.
Returning to the original problem, we can now integrate #int\ 1-1/(2x^2+2)-1/(4x+4)+1/(4x-4)\ dx#.
The first, third, and fourth can be easily solved using either the constant rule or u-substitution: #int\ dx=x#, #-int\ dx/(4x+4)=-(ln|x+1|)/4#, and #int\ dx/(4x-4)=(ln|x-1|)/4#.
The second one, #-int\ dx/(2x^2+2)=-1/2int\ dx/(x^2+1)#, requires knowing the identity #int\ dx/(x^2+1)=arctan(x)#.
Thus, we can combine everything to get the final answer: #x+1/4ln|(x-1)/(x+1)|-arctan(x)/2+C#.
