Question

∫(x^4+1)/x*(x^2+1)^2 dx 求不定积分

Answer 1

答案：Ln|x|+1/(x²+1)+C

分析：分子=(x²+1)²－2x²，故而原被积分式化为1/x－2x/(x²+1)²
前式的结果为Ln|x|
后式要作变化：－2xdx/(x²+1)²=－d(x²+1)/(x²+1)²，积分后1/(x²+1)

Answer 2

∫(x^4+1)/[x*(x^2+1)^2] dx
let
(x^4+1)/[x*(x^2+1)^2]≡ A/x + (B1x+B2)/(x^2+1) + (C1x+C2)/(x^2+1)^2
=>
(x^4+1)≡ A(x^2+1)^2 + (B1x+B2)x(x^2+1) + (C1x+C2)x
put x=0 , A=1

coef. of x^4
A+B1=1
B1= 0

coef. of x^3
B2 =0

coef. of x => C2 =0
coef. of x^2
2A+C1 =0
C1 =-2

(x^4+1)/[x*(x^2+1)^2]≡ 1/x - 2x/(x^2+1)^2
∫(x^4+1)/[x*(x^2+1)^2] dx
=∫[1/x - 2x/(x^2+1)^2] dx
=lnx - ∫ [2x/(x^2+1)^2] dx
= lnx + 1/(x^2+1) + C