Question

不定积分,dx/x^3+x^5

Answer 1

∫dx/(x^3+x^5)
=∫dx/[x^3(1+x^2)]
let
1/[x^3(1+x^2)]≡ A1/x + A2/x^2+A3/x^3 + (B1x+B2)/(1+x^2)
=>
1≡ A1x^2.(1+x^2) + A2x(1+x^2)+A3(1+x^2)+ (B1x+B2)x^3
x=0, =>A3=1

coef. of x =>A2=0

coef. of x^2
A1+A3 = 0
A1 =-A3 = -1

coef. of x^4
A1+B1=0
B1=-A1=1

coef. of x^3
A2+B2=0
B2=-A2 =0

1/[x^3(1+x^2)]≡ -1/x +1/x^3 + x/(1+x^2)

∫dx/(x^3+x^5)
=∫[-1/x +1/x^3 + x/(1+x^2)]dx
=-ln|x| -1/(3x^2) + (1/2)∫2x/(1+x^2)]dx
=-ln|x| -1/(3x^2) + (1/2)ln|1+x^2| + C

Answer 2

Answer 3

令t=(x^3+1)^(1/4) x=(t^4-1)^(1/3) dx=(4/3)*(t^4-1)^(-2/3)*t^2
原式=∫(t^4-1)^(5/3)*(4/3)*(t^4-1)^(-2/3)*t*dt
=4/3*∫(t^5-t)dt
=4/3*[(t^6)/6-(t^2)/2]+C
=(2t^6)/9-(2t^2)/3+C
=2/9*(x^3+1)^(3/2)-2/3*(x^3+1)^(1/2)+C

Answer 4

令t=1/x,则x=1/t
原式=∫1/（1/t^3+1/t^5）d(1/t)
=-∫t^3/(1+t^2)dt
=-∫[t(1+t^2)-t]/(1+t^2)dt
=-∫[t-t/(1+t^2)]dt
=-(1/2)t^2+(1/2)ln(1+t^2)+c
=1/2[ln(1+t^2)-t^2]+c
将t代回x
原式=1/2[ln(1+1/x^2)-1/x^2]+c

Answer 5