∫(x^5+x^4-8）／x^3-x dx求不定积分具体过程

∫(x^5+x^4-8）／(x^3-x)dx
=S(x^5-x^3+x^4-x^2+x^3-x+x^2-1+x-1-6)(x^3-x)dx
=S(x^2+x+1+1/x+1/x(x+1)-6/(x^3-x))dx
=1/3*x^3+1/2*x^2+x+lnx+ln(x/(1+x))-6S1/(x^3-x)dx
=1/3*x^3+1/2*x^2+x+lnx+ln(x/(1+x)-3S(1/(x-1)-2/x+1/(x+1))dx
=1/3*x^3+1/2*x^2+x+lnx+ln(x/(1+x)-3ln(x-1)+6lnx-3ln(x+1)+c
=1/3*x^3+1/2*x^2+x+8lnx-4ln(1+x)-3ln(x-1)+c设1/(x^3-x)=1/(x(x-1)(x+1))=A/(x-1)+B/x+C/(x+1)右边通分后分子为：A（x^2+x)+B(x^2-1)+C(x^2-x)=(A+B+C)x^2+(A-C)x-B左边分子为：1所以：A+B+C=0A-C=0-B=1B=-1A=C=1/2所以：S1/(x^3-x)dx=S1/2*(1/(x-1)-2/x+1/(x+1))dx=1/2*(ln(x-1)-2lnx+ln(x+1))+c