不定積分1 1 x 3 dx有什麼好方法

時間 2021-09-02 08:10:11

1樓:隱綠柳邸賦

-1/(x-1)(x²+x+1)

設=a/(x-1)+(bx+c)/(x²+x+1)通分後計算分母得1,所以

a(x²+x+1)+(bx+c)(x-1)=1(a+b)x²+(a-b+c)x+a-c=1a+b=0

a-b+c=0

a-c=1

解得a=1/3,b=-1/3,c=-2/3原式=a/(x-1)+(bx+c)/(x²+x+1)=[1/(x-1)-(x+2)/(x²+x+1)]/3∫[1/(x-1)-(x+0.5+1.5)/(x²+x+1)]/3dx=∫

[1/(x-1)-(x+0.5)/(x²+x+1)-1.5/(x²+x+1)]/3

dx=(1/3)[ln|x-1|-0.5ln(x²+x+1)]-∫0.5/[(x+1/2)²+3/4]dx重點解決

∫0.5/[(x+1/2)²+3/4]dx設(x+1/2)=[(根號3)/2]tantdx=[(根號3)/2]sec²t

dt∫0.5/[(x+1/2)²+3/4]dx=0.5

∫[1/(3/4sec²t)][(根號3)/2]sec²tdt=0.5*4*(根號3)/(3*2)∫1dt=(根號3/3)t+c

tant=(2x+1)/(根號3)

t=arctan[(2x+1)/(根號3)]∫0.5/[(x+1/2)²+3/4]dx=((根號3)/3)*arctan[(2x+1)/(根號3)]+c帶回

(1/3)[ln|x-1|-0.5ln(x²+x+1)]-∫0.5/[(x+1/2)²+3/4]dx=(1/3)[ln|x-1|-0.

5ln(x²+x+1)]-((根號3)/3)*arctan[(2x+1)/(根號3)]+c

2樓:犁玉蘭翠燕

沒什麼好辦法,對1/(1-x^3)分解得:

1/(1-x^3)=1/3(1-x)+(x+2)/3(1+x+x^2)

所以原式=∫1/3(1-x)dx+∫(x+2)/3(1+x+x^2)dx

=-ln

|x-1

|/3+ln(x^2+x+1)/6+2arctan[(2x+1)/根號3]/根號3+c

不定積分:∫ 1/(1-x^3) dx 有什麼好方法

3樓:西江樓望月

-1/(x-1)(x²+x+1)

設=a/(x-1)+(bx+c)/(x²+x+1)

通分後計算分母得1,所以

a(x²+x+1)+(bx+c)(x-1)=1

(a+b)x²+(a-b+c)x+a-c=1

a+b=0

a-b+c=0

a-c=1

解得a=1/3,b=-1/3,c=-2/3

原式=a/(x-1)+(bx+c)/(x²+x+1)

=[1/(x-1)-(x+2)/(x²+x+1)]/3

∫ [1/(x-1)-(x+0.5+1.5)/(x²+x+1)]/3 dx

=∫ [1/(x-1)-(x+0.5)/(x²+x+1)-1.5/(x²+x+1)]/3 dx

=(1/3)[ln|x-1|-0.5ln(x²+x+1)] - ∫0.5/[(x+1/2)²+3/4]dx

重點解決

∫0.5/[(x+1/2)²+3/4]dx

設(x+1/2)=[(根號3)/2]tant

dx=[(根號3)/2]sec²t dt

∫0.5/[(x+1/2)²+3/4]dx

=0.5 ∫[1/(3/4sec²t)][(根號3)/2]sec²t dt

=0.5*4*(根號3)/(3*2)∫1 dt

=(根號3/3)t+c

tant=(2x+1)/(根號3)

t=arctan[(2x+1)/(根號3)]

∫0.5/[(x+1/2)²+3/4]dx

=((根號3)/3)*arctan[(2x+1)/(根號3)] +c

帶回(1/3)[ln|x-1|-0.5ln(x²+x+1)] - ∫0.5/[(x+1/2)²+3/4]dx

=(1/3)[ln|x-1|-0.5ln(x²+x+1)] - ((根號3)/3)*arctan[(2x+1)/(根號3)] +c

4樓:

沒什麼好辦法,對1/(1-x^3)分解得:

1/(1-x^3)=1/3(1-x)+(x+2)/3(1+x+x^2)

所以原式=∫1/3(1-x)dx+∫(x+2)/3(1+x+x^2)dx

=-ln | x-1 |/3+ln(x^2+x+1)/6+2arctan[(2x+1)/根號3]/根號3+c

求不定積分∫[1/(1+x^3)]dx 要步驟

5樓:留秀雲建鳥

^||1+x^3=(x+1)(x^2-x+1)

用待定係數法:a/(x+1)+(bx+c)/(x^2-x+1)=1/(x+1)(x^2-x+1)

得a=1/3,b=-1/3,c=2/3

所以∫[1/(1+x^3)]dx

=1/3∫(1/(x+1))dx-1/3∫((x-2)/(x^2-x+1))dx

其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c

因為d(x^2-x+1)=(2x-1)dx,所以x-2=1/2(2x-1)-3/2

∫((x-2)/(x^2-x+1))dx=1/2∫(d(x^2-x+1)/(x^2-x+1))-3/2∫(1/(x^2-x+1))dx

其中∫(d(x^2-x+1)/(x^2-x+1))=ln|x^2-x+1|+c

∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根號3/2)^2))

因為∫(dx/(x^2+a^2))=(1/a)arctan(x/a)

所以∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根號3/2)^2))

=(2/根號3)arctan((x-1/2)/(根號3/2))+c

在乘上係數,整理∫[1/(1+x^3)]dx=1/3ln|x+1|-1/6|x^2-x+1|+(1/根號3)arctan((2x-1)/根號3)+c

6樓:童雲德慶戌

^∫(1-x)/(1+x^3)dx

這個就需要用因式分解

1+x^3=(1+x)(x^2-x+1)

將(1-x)化成這兩個因式的加和

(1-x)=(2/3)(x^2-x+1)-(1/3)(2x-1)(x+1)

∫(1-x)/(1+x^3)dx

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

dx=(2/3)∫1/(x+1)dx

-(1/3)

∫[(2x^2-2x+2)+(3x-3)]/(x^2-x+1)

dx=(2/3)

ln(x+1)-(2/3)x+(1/2)∫1/(x^2-x+1)d(x^2-x+1)+

(√3/3)arctan[(2x-1)/√3]

=(2/3)

lnix+1i-(2/3)x+(1/2)lnix^2-x+1i+(√3/3)arctan[(2x-1)/√3]+c

解答完畢,請指教,真麻煩啊呀

求不定積分∫[1/(1+x^3)]dx 要步驟

7樓:匿名使用者

|1+x^3=(x+1)(x^2-x+1)

用待定係數法:a/(x+1)+(bx+c)/(x^2-x+1)=1/(x+1)(x^2-x+1)

得a=1/3,b=-1/3,c=2/3

所以∫[1/(1+x^3)]dx =1/3∫(1/(x+1))dx-1/3∫((x-2)/(x^2-x+1))dx

其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c

因為d(x^2-x+1)=(2x-1)dx,所以x-2=1/2(2x-1)-3/2

∫((x-2)/(x^2-x+1))dx=1/2∫(d(x^2-x+1)/(x^2-x+1))-3/2∫(1/(x^2-x+1))dx

其中∫(d(x^2-x+1)/(x^2-x+1))=ln|x^2-x+1|+c

∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根號3/2)^2))

因為∫(dx/(x^2+a^2))=(1/a)arctan(x/a)

所以∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根號3/2)^2))

=(2/根號3)arctan((x-1/2)/(根號3/2))+c

在乘上係數,整理∫[1/(1+x^3)]dx=1/3ln|x+1|-1/6|x^2-x+1|+(1/根號3)arctan((2x-1)/根號3)+c

1/(x^3+x+1)的不定積分怎麼算啊,急需 200

8樓:不是苦瓜是什麼

1/(1+x^3)的不定積分求法如下:

1+x^3=(x+1)(x^2-x+1)

用待定係數法e68a8462616964757a686964616f31333431353262:a/(x+1)+(bx+c)/(x^2-x+1)=1/(x+1)(x^2-x+1)

得a=1/3,b=-1/3,c=2/3

所以∫[1/(1+x^3)]dx =1/3∫(1/(x+1))dx-1/3∫((x-2)/(x^2-x+1))dx

其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c

因為d(x^2-x+1)=(2x-1)dx,所以x-2=1/2(2x-1)-3/2

∫((x-2)/(x^2-x+1))dx=1/2∫(d(x^2-x+1)/(x^2-x+1))-3/2∫(1/(x^2-x+1))dx

其中∫(d(x^2-x+1)/(x^2-x+1))=ln|x^2-x+1|+c

∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根號3/2)^2))

因為∫(dx/(x^2+a^2))=(1/a)arctan(x/a)

所以∫(1/(x^2-x+1))dx=∫(dx/((x-1/2)^2+(根號3/2)^2))

=(2/根號3)arctan((x-1/2)/(根號3/2))+c

不定積分的公式

1、∫ a dx = ax + c,a和c都是常數

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + c,其中a為常數且 a ≠ -1

3、∫ 1/x dx = ln|x| + c

4、∫ a^x dx = (1/lna)a^x + c,其中a > 0 且 a ≠ 1

5、∫ e^x dx = e^x + c

6、∫ cosx dx = sinx + c

7、∫ sinx dx = - cosx + c

8、∫ cotx dx = ln|sinx| + c = - ln|cscx| + c

9、∫ tanx dx = - ln|cosx| + c = ln|secx| + c

10、∫ secx dx =ln|cot(x/2)| + c = (1/2)ln|(1 + sinx)/(1 - sinx)| + c = - ln|secx - tanx| + c = ln|secx + tanx| + c

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