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为-1/x，(1/2)/(x+1)，(1/2)/(x-1)，
(x^5+2x^2+1)/(x^3-x)可用分式除法，也可以配成x^5-x^3+x^3-x+x+2x^2+1形式。

Answer 2

(x^5+2x^2+1)/(x^3-x)=(x^4)/(x-1)+(2/3)(1/(x-1)+1/(x+1))-1/x
=[(x-1)(x^3+x^2+x+1)+1]/x+(2/3)(1/(x-1)+1/(x+1))-1/x
∫(x^5+2x^2+1)/(x^3-x)dx=∫{[(x-1)(x^3+x^2+x+1)+1]/x+(2/3)(1/(x-1)+1/(x+1))-1/x } dx
=(1/4)x^4+(1/3)x^3+(1/2)x^2+x+(3/2)Lnlx+1l+(5/2)Lnlx-1l-Lnlxl+c

Answer 3

答案来了