Question

怎样求不定积分:∫(x^5+2x^2+1)/(x^3-x)dx

Answer 1

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)|+(5/3)*
∫[(-1/x)+(1/2)/(x+1)+(1/2)/(x-1)]
=x^3/3+x+(2/3)ln|x^3-x|+(1/2)ln|(x-1)/(x+1)|-(5/3)ln|x|+(5/6)ln|x+1|+(5/6)ln|x-1|+C
=
x^3/3+x+(2/3)ln|x^3-x|+(4/3)ln|(x-1)+(1/3)ln|x+1)|-(5/3)ln|x|+C
注:1/[x(x+1)(x-1)用待定系数法得出系数-1/x(1/2)/(x+1)(1/2)/(x-1)
(x^5+2x^2+1)/(x^3-x)用分式除法也配成x^5-x^3+x^3-x+x+2x^2+1形式