Question

高等数学不定积分求解

求解。

Answer 1

=积分【x^2+x+1+(x^2+x+8)/(x^3-x)】dx
=x^3/3+x^2/2+x+积分[（x^2-1）+x+9]/[x(x^2-1)]dx
=x^3/3+x^2/2+x+积分[1/x+1/(x^2-1)+9/(x^2-1)*1/x]dx
=x^3/3=x^2/2+x+lnx+积分1/2*[1/(x-1)-1/(x+1)]dx+积分[9/(x^2-1)*1/2*1/x^2]d(x^2)
=x^3/3+x^2/2+x+lnx+1/2*ln[(x-1)/(x+1)]+9/2*ln[(x^2-1)/x^2]+C
=x^3/3+x^2/2+x+lnx+1/2*ln[(x-1)-ln(x+1)]+9/2*ln[(x-1)+9/2*ln(x+1)-9lnx+C
=x^3/3+x^2/2+x-8lnx+5*ln(x-1)+4*ln(x+1)+C

Answer 2

题目中 的数 看不清。 是否 5，1
3 ，

追问

5,4，
3,1

追答

趣味 和 启发 式，好吗：
(x^5+x^4-8)/(x^3-x)
=[(x^5-x^3)+(x^3-x) +(x-1) + (x^4-x^2) +(x^2-1) +(-6) ] / (x^3-x)
= x^2 +1 +(x-1) / (x^3-x )+ x +(x^2-1)/ (x^3-x) ) +(-6) / (x^3-x)
= x^2 +1 + 1/[x(x+1)] +x 1/x +(-6) / (x^3-x)
= x^2 +x +1 + 1/[x(x+1)] + 1/x +(-6) / (x^3-x)
= x^2 +x +1 +[1/x - 1/(x+1)] + 1/x +(-6) / [x(x^2-1]
=x^2 +x +1 +2/x - 1/(x+1) + (-6x) / [x^2(x^2-1]
=x^2 +x +1 +2/x - 1/(x+1) -3(2x)/ [x^2(x^2-1]
={x^2 +x +1 +2/x - 1/(x+1) } + 3(2x)[ 1/x^2 - 1/(x^2-1) ]
={x^2 +x +1 +2/x - 1/(x+1) } + 3(2x)/x^2 - 3(2x)/(x^2-1)

2xdx=d(x^2) 2xdx=d(x^2-1)

积分【(x^5+x^4-8)/(x^3-x)】dx
=积分【 {x^2 +x +1 +2/x - 1/(x+1) } + 3(2x)/x^2 - 3(2x)/(x^2-1) 】dx
=积分【x^2 +x +1 +2/x - 1/(x+1) 】dx+ 积分【[ 3/x^2 】d(x^2) - 积分【 3/(x^2-1) 】d（x^2-1)
=.x^3/3+x^2/2+x+......

Answer 3

请把题目补充清楚。