Let us first obtain partial fractions of #(2x-2)/((x^4-1)(x-3))# and as #x^4-1=(x+1)(x-1)(x^2+1)#, these will be
#(2x-2)/((x^4-1)(x-3))=(2(x-1))/((x+1)(x-1)(x^2+1)(x-3))=2/((x+1)(x^2+1)(x-3))#
and this can be expressed as
#2/((x+1)(x^2+1)(x-3))=A/(x+1)+(Bx+C)/(x^2+1)+D/(x-3)#
#2=A(x-3)(x^2+1)+(Bx+C)(x+1)(x-3)+D(x+1)(x^2+1)#
putting #x=-1# and #x=3# in this we get
#2=-8A# i.e. #A=-1/4# and #40D=2# i.e. #D=1/20#
and comparing coefficients of #x^3# and constant term,
we have, #0=A+B+D# i.e. #B=-A-D=1/4-1/20=4/20=1/5#
and #2=-3A-3C+D# or #3C=-2-3(-1/4)+1/20=-2+3/4+1/20=(-24)/20=-6/5# and #C=-2/5# and
#2/((x+1)(x^2+1)(x-3))=-1/(4(x+1))+(x-2)/(5(x^2+1))+1/(20(x-3))#
and #int(2x-2)/((x^4-1)(x-3))dx#
= #int-1/(4(x+1))dx+int(x-2)/(5(x^2+1))dx+int1/(20(x-3))dx#
= #int-1/(4(x+1))dx+1/5(intx/(x^2+1)dx-int2/(x^2+1)dx)+int1/(20(x-3))dx#
= #-1/4ln|x+1|+1/5(1/2ln(x^2+1)-2tan^(-1)x)+1/20ln|x-3|#
= #-1/4ln|x+1|+1/10ln(x^2+1)-2/5tan^(-1)x+1/20ln|x-3|#
