怎样求不定积分：∫(x^5+2x^2+1)/(x^3-x)dx

原式=∫(x^2+1)dx+∫(2x^2+x+1)dx/(x^3-x) =x^3/3+x+(2/3) ∫d(x^3-x)/（x^3-x）+∫(x+5/3)dx/(x^3-x) =x^3/3+x+(2/3)ln|x^3-x|+∫dx/(x^2-1)+(5/3) ∫dx/[x(x+1)(x-1) =x^3/3+x+(2/3)ln|x^3-x|+(1/2)ln|(x-1)/(x+1)|+