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|
