问几个不定积分，怎么解合适那、、

Question

∫x^2/(x^2+1)dx

Answer 1

1. ∫dx/(x-2)²(x-3)=∫dx/(x-2)(x-3)-∫dx/(x-2)²=∫dx/(x-3)-∫dx/(x-2)-∫dx/(x-2)²=ln│x-3│-ln│x-2│+1/(x-2)+C
=ln│(x-3)/(x-2)│+1/(x-2)+C
2. ∫sinxsin2xsin3xdx=∫sinx·2sinxcosx·(3sinx-4sin³x)dx=∫[6sin³x-8(sinx)^5]dsinx
=(6/4)(sinx)^4-(8/6)(sinx)^6+C
=(3/2)(sinx)^4-(4/3)(sinx)^6+C
3. ∫(x²-5x+9)dx/(x²-5x+6)=∫1dx+3∫dx/(x-2)(x-3)=x+3∫dx/(x-3)-3∫dx/(x-2)=x+3ln│(x-3)/(x-2)│+C
4. ∫cosxdx/sinx(1+sinx)²=∫dsinx[1/sinx(1+sinx)-1/(1+sinx)²]=∫dsinx[1/sinx-1/(1+sinx)-1/(1+sinx)²]
=ln│sinx/(1+sinx)│+1/(1+sinx)+C
5. x=tan(a),a=arctan(x); dx=(sec(a))^2da, 1+x^2=1+(tan(a))^2=(sec(a))^2

∫x²dx/(x²+1)=∫dx-∫dx/(x²+1)=∫dx-∫da (sec(a))^2/(sec(a))^2)=x-a+C
=x-arctanx+C

Answer 2

答案在图片上

 

累死我了请加分

追问

分可以加，但是你的3题不对额

追答

不好意思,写错了！现在更正过来
∫(x²-5x+9)dx/(x²-5x+6)=∫1dx+3∫dx/(x-2)(x-3)=x+3∫dx/(x-3)-3∫dx/(x-2)=x+3ln|(x-2)/(x-3)|+C

Answer 3

(1)
let
1/[(x-2)^2(x-3)] = a1/(x-2) + a2/(x-2)^2 + b/(x-3)
=>
1= a1(x-2)(x-3) +a2(x-3) +b(x-2)^2
put x=2 , a2=-1
put x=3, b=1
coef. of x^3
a1+a2+b =0
a1=0
1/[(x-2)^2(x-3)] = -1/(x-2)^2 + 1/(x-3)
∫1/[(x-2)^2(x-3)] dx
=∫ [-1/(x-2)^2 + 1/(x-3)] dx
= 1/(x-2) +ln|x-3| + C
(2)
∫ sinx sin2xsin3x dx
=(1/2)∫ (cosx -cos3x)sin3x dx
= (1/2)∫ sin3xcosx dx - (1/4)∫ sin6x dx
= (1/4)∫ (sin4x - sin2x) dx + (1/24)cos6x
= -(1/16)cos4x + (1/8)cos2x+ (1/24)cos6x + C
(3)
∫(x^2-5x+9)/(x^2-5x+6) dx
=∫ [1+ 3/(x^2-5x+6) ]dx
= x+ 3∫ 1/(x^2-5x+6) dx
consider
x^2-5x+6 = (x-5/2)^2 -(1/2)^2
let
x-5/2 = (1/2)secy
dx = (1/2)secytany dy
∫ 1/(x^2-5x+6) dx
=∫ [4/(tany)^2](1/2)secytany dy
=2∫ secy/tany dy
=2∫ cscy dy
=2 ln|cscy-coty| + C'
= 2ln|(x-3)/√(x^2-5x+6)| + C'
∫(x^2-5x+9)/(x^2-5x+6) dx
= x+ 3∫ 1/(x^2-5x+6) dx
= x+6ln|(x-3)/√(x^2-5x+6)| + C
(4)
∫ cosx/[sinx(1+sinx)^2] dx
let
1/[sinx(1+sinx)^2] = a1/(1+sinx) +a2/(1+sinx)^2 + b/sinx
1= a1(1+sinx)sinx +a2sinx + b(1+sinx)^2
sinx=-1 , a2 =-1
sinx=0, b=1
coef. of sinx
a1+a2+2b =0
a1=-1
1/[sinx(1+sinx)^2] = -1/(1+sinx) -1/(1+sinx)^2 + 1/sinx
∫ cosx/[sinx(1+sinx)^2] dx
=∫ cosx[-1/(1+sinx) -1/(1+sinx)^2 + 1/sinx] dx
= ln|sinx/(1+sinx)| + 1/(1+sinx) + C
(5)
∫x^2/(x^2+1)dx
=∫[1-1/(x^2+1)]dx
= x - arctanx + C

追问

1 3不对啊
尽量用中文，不是我看不懂，是和数字搭配好晕啊，谢谢

追答

(1)
设
1/[(x-2)^2(x-3)] = a1/(x-2) + a2/(x-2)^2 + b/(x-3)
=>
1= a1(x-2)(x-3) +a2(x-3) +b(x-2)^2
令 x=2 , a2=-1
令 x=3, b=1
x^2 项数
a1+b =0
a1=-1
1/[(x-2)^2(x-3)] = -1/(x-2)-1/(x-2)^2 + 1/(x-3)
∫1/[(x-2)^2(x-3)] dx
=∫ [-1/(x-2) -1/(x-2)^2 + 1/(x-3)] dx
= 1/(x-2) +ln|(x-3)/(x-2)| + C

(3)
∫(x^2-5x+9)/(x^2-5x+6) dx 有没有错？？？？？？？

追问

嗯 3题没有错，答案是x+3In|(x-3)/(x-2)|+c

追答

∫(x^2-5x+9)/(x^2-5x+6) dx
=∫ [1+ 3/(x^2-5x+6) ]dx
=∫ (1 + 3/[(x-2)(x-3)] )dx
= x + 3∫ [1/(x-2) -1/(x-3)]dx
= x+ ln|(x-2)/(x-3)| + C

Answer 4

Answer 5

很难の说 让偶这个数学专业的学生情何以堪。

追问

我也是数学专业的，不知道怎么解啊

追答

以前学的 忘得差不多叻 待会儿看看书去=-=