Question

求不定积分谢谢

Answer 1

解析：
习惯了用计算器
详见附图

Answer 2

let
1/[x^4.(1+x^2)]≡ A/x+B/x^2+C/x^3+D/x^4 + (Ex+F)/(1+x^2)
=>
1≡ Ax^3.(1+x^2)+Bx^2.(1+x^2)+Cx.(1+x^2)+D(1+x^2) + (Ex+F)x^4
x=0, =>D =1
coef. of x^2 : B+D=0 => B=-1
coef. of x^4: B+F=0 => F=1
coef. of x : => C=0
coef. of x^3: C+A=0 =>A=0
coef. of x^5 : A+E=0 => E=0
1/[x^4.(1+x^2)]≡ -1/x^2+1/x^4 + 1/(1+x^2)
∫dx/[x^4.(1+x^2)]
=∫ {-1/x^2+1/x^4 + 1/(1+x^2) } dx
= 1/x - 1/(3x^3) + arctanx + C

Answer 3

问题不全没法求

Answer 4

∫dx/[x⁴(1+x²)]
=∫[1/x⁴-1/x²+1/(1+x²)]dx
=-1/(3x³)+ 1/x +arctanx +C

追问

也是怎么拆开的能详细一点吗

追答

要用到待定系数法。
令1/(1+x²) -(x²+a)/x⁴=1/[x⁴(1+x²)]
x⁴-(1+x²)(x²+a)=1
(a+1)x²+(a+1)=0
(a+1)(x²+1)=0
只有a=-1
因此
1/[x⁴(1+x²)]
=1/(1+x²) -(x²-1)/x⁴
=1/(1+x²) -1/x² +1/x⁴
数学计算的基本功。如果你连基本功都不具备，那就悲催了，知道方法也没用，也解不出来。