设1/[x^2*(x+1)]=a/(x+1)+(bx+c)/x^2,则1=ax^2+(x+1)(bx+c)=(a+b)x^2+(b+c)x+c,所以a+b=0=b+c,c=1,解得b=-1,a=1.所以∫dx/[x^2*(x+1)]=∫[1/(x+1)-1/x+1/x^2]dx=ln|x+1|-ln|x|-1/x+c.
