Let us first obtain partial fractions of (2x-2)/((x^4-1)(x-3))2x−2(x4−1)(x−3) and as x^4-1=(x+1)(x-1)(x^2+1)x4−1=(x+1)(x−1)(x2+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))2x−2(x4−1)(x−3)=2(x−1)(x+1)(x−1)(x2+1)(x−3)=2(x+1)(x2+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(x+1)(x2+1)(x−3)=Ax+1+Bx+Cx2+1+Dx−3
2=A(x-3)(x^2+1)+(Bx+C)(x+1)(x-3)+D(x+1)(x^2+1)2=A(x−3)(x2+1)+(Bx+C)(x+1)(x−3)+D(x+1)(x2+1)
putting x=-1x=−1 and x=3x=3 in this we get
2=-8A2=−8A i.e. A=-1/4A=−14 and 40D=240D=2 i.e. D=1/20D=120
and comparing coefficients of x^3x3 and constant term,
we have, 0=A+B+D0=A+B+D i.e. B=-A-D=1/4-1/20=4/20=1/5B=−A−D=14−120=420=15
and 2=-3A-3C+D2=−3A−3C+D or 3C=-2-3(-1/4)+1/20=-2+3/4+1/20=(-24)/20=-6/53C=−2−3(−14)+120=−2+34+120=−2420=−65 and C=-2/5C=−25 and
2/((x+1)(x^2+1)(x-3))=-1/(4(x+1))+(x-2)/(5(x^2+1))+1/(20(x-3))2(x+1)(x2+1)(x−3)=−14(x+1)+x−25(x2+1)+120(x−3)
and int(2x-2)/((x^4-1)(x-3))dx∫2x−2(x4−1)(x−3)dx
= int-1/(4(x+1))dx+int(x-2)/(5(x^2+1))dx+int1/(20(x-3))dx∫−14(x+1)dx+∫x−25(x2+1)dx+∫120(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∫−14(x+1)dx+15(∫xx2+1dx−∫2x2+1dx)+∫120(x−3)dx
= -1/4ln|x+1|+1/5(1/2ln(x^2+1)-2tan^(-1)x)+1/20ln|x-3|−14ln|x+1|+15(12ln(x2+1)−2tan−1x)+120ln|x−3|
= -1/4ln|x+1|+1/10ln(x^2+1)-2/5tan^(-1)x+1/20ln|x-3|−14ln|x+1|+110ln(x2+1)−25tan−1x+120ln|x−3|
