Question

1/x^8*(1+x^2)的不定积分

是1/（x^8*(1+x^2)）

Answer 1

若是∫(1+x²)/x^8 dx
= ∫(x^-8 + x^-6) dx
= x^(-8+1) / (-8+1) + x^(-6+1) / (-6+1) + C
= -1/[7x^7] - 1/[5x^5]
= -[7x²+5] / [35x^7] + C

若是∫dx/[x^8(1+x²)]
令1/[x^8*(1+x²)] = A/x^8 + B/x^6 + C/x^4 + D/x^2 + E/(x²+1)
待定系数法，过程太烦，召唤答案~~
A = 1，B = -1，C = 1，D = -1，E = 1
原式= ∫[1/x^8 - 1/x^6 + 1/x^4 - 1/x^2 + 1/(x²+1)] dx
= -1/[7x^7] + 1/[5x^5] + 1/[3x^3] + 1/x + arctanx + C

Answer 2

=X^(-8)+X^(-6)
不定积分：-8X^(-9)-6X^(-7)+C