Perform the decomposition into partial fractions
(x^3-x^2+1)/(x^4-x^3)=(x^3-x^2+1)/(x^3(x-1))x3−x2+1x4−x3=x3−x2+1x3(x−1)
=A/(x^3)+B/(x^2)+C/(x)+D/(x-1)=Ax3+Bx2+Cx+Dx−1
=(A(x-1)+B(x(x-1))+C(x^2(x-1))+D(x^3))/((x^4-x^3))=A(x−1)+B(x(x−1))+C(x2(x−1))+D(x3)(x4−x3)
The denominators are the same, compare the numerators
(x^3-x^2+1)=A(x-1)+B(x(x-1))+C(x^2(x-1))+D(x^3)(x3−x2+1)=A(x−1)+B(x(x−1))+C(x2(x−1))+D(x3)
Let x=0x=0, =>⇒, 1=-A1=−A, =>⇒, A=-1A=−1
Coefficients of xx
0=A-B0=A−B, =>⇒, B=A=-1B=A=−1
Coefficients of x^2x2
-1=B-C−1=B−C, =>⇒, C=B+1=-1+1=0C=B+1=−1+1=0
Coefficients of x^3x3
1=C+D1=C+D, D=1-C=1-0=1D=1−C=1−0=1
Therefore,
(x^3-x^2+1)/(x^4-x^3)=-1/(x^3)-1/(x^2)+0/(x)+1/(x-1)x3−x2+1x4−x3=−1x3−1x2+0x+1x−1
So,
int((x^3-x^2+1)dx)x^4-x^3)=-int(1dx)/(x^3)-int(1dx)/(x^2)+int(1dx)/(x-1)∫((x3−x2+1)dx)x4−x3)=−∫1dxx3−∫1dxx2+∫1dxx−1
=1/(2x^2)+1/x+ln(|x-1|)+C=12x2+1x+ln(|x−1|)+C
