Question

图上两题求不定积分

Answer 1

【第5题分母上的5次根号里面的x是几次方？看不清楚，没法帮你作。】

Answer 2

let
x/[(x+1)(x+2)(x+3)]≡ A/(x+1)+B/(x+2)+C/(x+3)
=>
x≡ A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)

x=-1
-1=2A
A= -1/2

x=-2
-2=-B
B=2

x=-3
-3=2C
C=-3/2

x/[(x+1)(x+2)(x+3)]≡-(1/2)[1/(x+1)]+2[1/(x+2)]-(2/3)[1/(x+3)]
∫x/[(x+1)(x+2)(x+3)] dx
=∫ { -(1/2)[1/(x+1)]+2[1/(x+2)]-(2/3)[1/(x+3)] } dx
=-(1/2)ln|x+1|+2ln|x+2|-(3/2)ln|x+3| + C

(2)
∫dx/[x(x^(1/2)+x^(2/5))]
let
x^(1/10) = u
(1/10)x^(-9/10) dx = du
dx = (10u^9) du

∫dx/[x(x^(1/2)+x^(2/5))]
=10∫du/[u(u^5+u^4)]
=10∫du/[u^5.(u+1)]
let
1/[u^5.(u+1)] ≡ A5/u^5 +A4/u^4+A3/u^3+A2/u^2+A1/u+ B/(u+1)
=>
1 ≡ A5(u+1) +A4u(u+1)+A3u^2.(u+1)+A2u^3.(u+1)+A1u^4.(u+1)+ Bu^5
u=-1, => B =-1
coef. of constant =>A5 = 1

coef. of u
A5+A4 =0
A4 = -1

coef. of u^2
A4+A3=0
A3 =1

coef. of u^3
A3+A2=0
A2=-1

coef. of u^4
A2+A1=0
A1=1

1/[u^5.(u+1)] ≡ 1/u^5 -1/u^4+1/u^3-1/u^2+1/u- 1/(u+1)
∫du/[u^5.(u+1)]
=∫ { 1/u^5 -1/u^4+1/u^3-1/u^2+1/u- 1/(u+1)} du
= -1/(4u^4) + 1/(3u^3) - 1/(2u^2) + ln|u/(u+1)| + C'

∫dx/[x(x^(1/2)+x^(2/5))]
=10∫du/[u^5.(u+1)]
=10 {-1/(4u^4) + 1/(3u^3) - 1/(2u^2) + ln|u/(u+1)| } + C
=10{-1/(4x^(2/5))+1/(3x^(3/10))-1/(2x^(1/5))+ln|x^(1/10)/(x^(1/10)+1)| }+C