按小数部分积分–积分

如果f(x)和g(x)是多项式函数，则该函数。 g(x)≠0，则f(x)/ g(x)称为有理函数。如果度f(x)<度g(x)，则f(x)/ g(x)被称为适当的有理函数。如果度f(x)<度g(x)，则f(x)/ g(x)称为不合理的有理函数。如果f(x)/ g(x)是不正确的有理函数，则通过将f(x)除以g(x)，我们可以将f(x)/ g(x)表示为多项式和适当有理数的和函数。

部分分数

任何适当的有理函数p(x)/ q(x)都可以表示为有理函数的总和，每个函数具有最简单的因子q(x)，每个这样的分数都被称为分函数，并且将获得这些函数的过程称为将给定函数分解或分解为部分分数。

分部积分

例如，假设我们要评估∫[p(x)/ q(x)] dx，其中p(x)/ q(x)在适当的有理分数内。在这样的情况下，在可以很容易地进行积分之后，通过使用部分分数分解，我们可以以更简单的有理函数之和的形式编写被积分数。在这里，可以根据问题获得A，B，C等的值。

Factors in the denominator or denominators in rational functions	Corresponding Partial Fractions
(x – a)	A/(x – a)
(x – b)2	A/(x – b) + A/(x – b)2
(x – c)3	A/(x – c) + B/(x – c)2 + C/(x – c)3
(ax2 + bx +c )	Ax + B/(ax2 + bx + c)

例子

示例1：评估∫(x – 1)/(x + 1)(x – 2)dx?

解决方案：

Let (x – 1)/(x + 1)(x – 2) = A/(x + 1) + B/(x – 2)

Then, (x – 1) = A(x – 2) + B(x + 1) ………………(i)

Putting x = -1 in (i), we get A = 2/3

Putting x = 2 in (i), we get B = 1/3

Therefore,  

(x – 1)/(x + 1)(x – 2) = 2/3(x + 1) + 1/3(x – 2)

=> ∫(x – 1)/(x + 1)(x – 2) = 2/3∫dx/(x + 1) + 1/3∫dx/(x – 2)

                                       = 2/3log | x + 1 | + 1/3 log | x – 2 | + C

示例2：评估∫dx/ x {6(log x) ² + 7log x + 2}?

解决方案：

Putting log x = t and 1/x dx = dt, we get

I = ∫dx/x{6(log x)² + 7log x + 2} = ∫dt/(6t² + 7t + 2) = ∫dt/(2t + 1)(3t + 2)

Let 1/(2t + 1)(3t + 2) = A/(2t + 1) + B/(3t + 2)

Then, 1 ≡ A(3t + 2) + B(2t + 1)  …………………….(i)

Putting t = -1/2 in (i), we get A = 2

Putting t = -2/3 in (i), we get B = -3

Therefore, 1/(2t + 1)(3t + 2) = 2/(2t + 1) + (-3)/(3t + 2)

=> I = ∫dt/(2t + 1)(3t + 2)

        = ∫2dt/(2t + 1) – ∫3dt/(3t – 2)

        = log | 2t + 1 | – log | 3t + 2 |

        = log | 2t + 1 |/log | 3t + 2 | + C

        = log | 2 log x + 1 | / log | 3 log x + 2 | + C

示例3：评估∫dx/(x ³ + x ² + x + 1)?

解决方案：

We have 1/(x³ + x² + x + 1) = 1/x²(x + 1) + (x + 1) = 1/(x + 1)(x² + 1)

Let 1/(x + 1)(x² + 1) = A/(x + 1) + Bx + C/(x² + 1)  ……………………(i)

=> 1 ≡ A(x² + 1) + (Bx + C) (x + 1)

Putting x = -1 on both sides of (i), we get A = 1/2.

Comparing coefficients of x² on the both sides of (i), we get

A + B = 0 => B = -A = -1/2

Comparing coefficients of x on the both sides of (i), we get

B + C = 0 => C = -B = 1/2

Therefore, 1/(x + 1) (x² + 1) = 1/2(x + 1) + (-1/2x + 1/2)/(x² + 1)

Therefore, ∫1/(x + 1) (x² + 1) = ∫dx/(x + 1) (x² + 1)

                                              = 1/2∫dx/(x + 1) – 1/2∫x/(x² + 1)dx + 1/2∫dx/(x² + 1)

                                              = 1/2∫dx/(x + 1) – 1/4∫2x/(x² + 1)dx + 1/2∫dx/(x² + 1)

                                              = 1/2 log | x + 1 | – 1/4 log | x² + 1 | + 1/2 tan^-1x + C

实施例4：评估^∫x2 /(X ² + 2)(X ² + 3)DX?

解决方案：

Let x²/(x² + 2) (x² + 3) = y/(y + 2)(y + 3) where x² = y.

Let y/(y + 2) (y + 3) = A/(y + 2) + B/(y + 3)

=> y ≡ A(y + 3) + B/(y + 2)   ………………(i)

Putting y = -2 on the both sides of (i), we get A = -2.

Putting y = -3 on the both sides of (i), we get B = 3.

Therefore, y/(y + 2) (y + 3) = -2/(y + 2) + 3/(y + 3)

=> x²/(x² + 2) (x² + 3) = -2/(x² + 2) + 3/(x² + 3)

=> ∫x²/(x² + 2) (x² + 3) = -2∫dx/(x² + 2) + 3∫dx/(x² + 3)

                                     = -2/√2tan^-1(x/√2) + 3/√3tan^-1(x/√3) + C

                                     = -√2tan^-1(x/√2) + √3tan^-1(x/√3) + C

范例5：评估∫dx/ x(x ⁴ +1)?

解决方案：

We have

I = ∫dx/x(x⁴ + 1) = ∫x³/x⁴ (x⁴ + 1) dx [multiplying numerator and denominator by x³].

Putting x⁴ = t and 4x³dx = dt, we get

I = 1/4∫dt/t(t + 1)

  = 1/4∫{1/t – 1/(t + 1)}dt   [by partial fraction]

  = 1/4∫1/t dt – 1/4∫1/(t + 1)dt

  = 1/4 log | t | – 1/4 log | t + 1 | + C

  = 1/4 log | x⁴ | – 1/4 log | x⁴ + 1 | + C

  = (1/4 * 4) log | x | – 1/4 log | x⁴ + 1 | + C

  = log | x | – 1/4 log | x⁴ + 1 | + C