Question

求∫1/x^3+1的不定积分

Answer 1

先分解因式：

∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx

= ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx

1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C

1 = (A + B)x² + (- A + B + C)x + (A + C)

{ A + B = 0

{ - A + B + C = 0

{ A + C = 1

(A + B) - (- A + B + C) = 0 ==> { 2A - C = 0

{A + C = 1

(2A - C) + (A + C) = 1 ==> 3A = 1 ==> A = 1/3

B = - 1/3,C = 2/3

原式 = (1/3)∫ dx/(x + 1) - (1/3)∫ (x - 2)/(x² - x + 1) dx

= (1/3)∫ d(x + 1)/(x + 1) - (1/3)∫ [(2x - 1)/2 - 3/2]/(x² - x + 1) dx

= (1/3)ln(x + 1) - (1/6)∫ d(x² - x + 1)/(x² - x +1) + (1/2)∫ d(x - 1/2)/[(x - 1/2)² + 3/4]

= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/2)(2/√3)arctan[(x - 1/2) * 2/√3] + C

= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/√3)arctan(2x/√3 - 1/√3) + C

扩展资料

不定积分的公式

1、∫ a dx = ax + C，a和C都是常数

2、∫ x^a dx = [x^(a + 1)]/(a + 1) + C，其中a为常数且 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

根据牛顿-莱布尼茨公式，许多函数的定积分的计算就可以简便地通过求不定积分来进行。这里要注意不定积分与定积分之间的关系：定积分是一个数，而不定积分是一个表达式，它们仅仅是数学上有一个计算关系。

一个函数，可以存在不定积分，而不存在定积分，也可以存在定积分，而没有不定积分。连续函数，一定存在定积分和不定积分；若在有限区间[a,b]上只有有限个间断点且函数有界，则定积分存在；若有跳跃、可去、无穷间断点，则原函数一定不存在，即不定积分一定不存在。

Answer 2

解答过程如下：

先分解因式：

∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx

= ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx

1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C

1 = (A + B)x² + (- A + B + C)x + (A + C)

{ A + B = 0

{ - A + B + C = 0

{ A + C = 1

(A + B) - (- A + B + C) = 0 ==> { 2A - C = 0

{A + C = 1

(2A - C) + (A + C) = 1 ==> 3A = 1 ==> A = 1/3

B = - 1/3,C = 2/3

原式 = (1/3)∫ dx/(x + 1) - (1/3)∫ (x - 2)/(x² - x + 1) dx

= (1/3)∫ d(x + 1)/(x + 1) - (1/3)∫ [(2x - 1)/2 - 3/2]/(x² - x + 1) dx

= (1/3)ln(x + 1) - (1/6)∫ d(x² - x + 1)/(x² - x +1) + (1/2)∫ d(x - 1/2)/[(x - 1/2)² + 3/4]

= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/2)(2/√3)arctan[(x - 1/2) * 2/√3] + C

= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/√3)arctan(2x/√3 - 1/√3) + C

扩展资料

分部积分：

(uv)'=u'v+uv'

得：u'v=(uv)'-uv'

两边积分得：∫ u'v dx=∫ (uv)' dx - ∫ uv' dx

即：∫ u'v dx = uv - ∫ uv' d,这就是分部积分公式

也可简写为：∫ v du = uv - ∫ u dv

常用积分公式：

1）∫0dx=c 

2）∫x^udx=(x^(u+1))/(u+1)+c

3）∫1/xdx=ln|x|+c

4）∫a^xdx=(a^x)/lna+c

5）∫e^xdx=e^x+c

6）∫sinxdx=-cosx+c

7）∫cosxdx=sinx+c

8）∫1/(cosx)^2dx=tanx+c

9）∫1/(sinx)^2dx=-cotx+c

10）∫1/√（1-x^2) dx=arcsinx+c

Answer 3

先分解因式：
∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx
= ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx
1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C
1 = (A + B)x² + (- A + B + C)x + (A + C)
{ A + B = 0
{ - A + B + C = 0
{ A + C = 1
(A + B) - (- A + B + C) = 0 ==> { 2A - C = 0
{A + C = 1
(2A - C) + (A + C) = 1 ==> 3A = 1 ==> A = 1/3
B = - 1/3，C = 2/3
原式 = (1/3)∫ dx/(x + 1) - (1/3)∫ (x - 2)/(x² - x + 1) dx
= (1/3)∫ d(x + 1)/(x + 1) - (1/3)∫ [(2x - 1)/2 - 3/2]/(x² - x + 1) dx
= (1/3)ln(x + 1) - (1/6)∫ d(x² - x + 1)/(x² - x +1) + (1/2)∫ d(x - 1/2)/[(x - 1/2)² + 3/4]
= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/2)(2/√3)arctan[(x - 1/2) * 2/√3] + C
= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/√3)arctan(2x/√3 - 1/√3) + C

满意请采纳，谢谢☆∧_∧☆

Answer 4

=∫1/x^3dx+∫dx=(-1/2)x^-2+x+c