(x^5 + 1)/(x^6 - x^4)=(x^5+1)/(x^4(x^2-1))=x5+1x6−x4=x5+1x4(x2−1)=
=c_1/x+c_2/x^2+c_3/x^3+c_4/x^4+c_5/(x+1)+c_6/(x-1)=c1x+c2x2+c3x3+c4x4+c5x+1+c6x−1
Then
int ((x^5 + 1)/(x^6 - x^4))dx=int(c_1/x+c_2/x^2+c_3/x^3+c_4/x^4+c_5/(x+1)+c_6/(x-1))dx=∫(x5+1x6−x4)dx=∫(c1x+c2x2+c3x3+c4x4+c5x+1+c6x−1)dx=
=c_1logabsx+c_2/x-c_3/(2x^2)-c_4/(3x^3)+c_4log abs(x+1)+c_5 log abs(x-1)+C=c1log|x|+c2x−c32x2−c43x3+c4log|x+1|+c5log|x−1|+C
The c_kck determination follows.
Equating
(x^5 + 1)/(x^6 - x^4)=c_1/x+c_2/x^2+c_3/x^3+c_4/x^4+c_5/(x+1)+c_6/(x-1) x5+1x6−x4=c1x+c2x2+c3x3+c4x4+c5x+1+c6x−1
The condition to mantain the equality forall x in RR is the observation of
{(1 + c_4=0), (c_3=0), (c_2 - c_4=0), (c_1 - c_3=0), (c_2 - c_5 + c_6=0), (c_1 + c_5 + c_6-1=0):}
Solving we obtain
(c_1 = 0, c_2= -1, c_3= 0, c_4 = -1, c_5 = 0, c_6= 1)
