This section introduces how to integrate a rational function (a ratio of polynomials) by expressing it as a sum of partial fractions, called partial fractions.

To see how the method of partial fractionsworks in general, consider a rational function

      f(x) = P(x)/Q(x)

where P(x) and Q(x) are polynomials. Assume the degree of P(x) is less than the degree of Q(X) (called proper rational function). To integration this proper rational function, first express it as a sum of partial fractions of the form

      A/(ax + b)ⁱ or (Ax + B)/(ax² + bx + c)^j

where A, B, a, b, and C are constant, and i, j are any positive integer. There are four cases that occur.

Case 1: The denominator Q(x) is a product of distinct linear factors

Since the denominator Q(x) is a product of distinct linear factors, it can be written as

      Q(x) = (a₁x + b₁)(a₂x + b₂) ... (a_kx + b_k)

where no factor is repeated. In this case the proper rational function can be expressed as

where A₁, A₂,..., A_k are constant. The following example shows how to determine these constants.

Consider a proper rational function

      P(x)/Q(x) = (2x + 3)/(x² - 1)

Since x² -1 = (x + 1)(x - 1), the partial fraction decomposition has the form

Multiplying both sides of the above equation by (x + 1)(x - 1) gives,

2x + 3 = A₁(x -1) + A₂(x + 1) = (A₁ + A₂)x + (A₂ - A₁)

The polynomials above are identical, so their coefficients must be equal.

      2 = A₁ + A₂
      3 = A₂ - A₁

Solving the above two equations gives A₁ and A₂.

      A₁ = -1/2 and A₂ = 5/2

Then the partial fraction of this proper rational function is

Recall, the integration of A_k /(a_kx + b_k) is

where C is a Constant.

Therefore, the integration of the proper rational function is

where C is a Constant.

Case 2: Q(x) is a product of linear factors, some of which are repeated.

Suppose the first linear factor (a₁x + b₁) is repeated t times, then the proper rational function can be expressed as

The constant A₁₁, A₁₂,..., A_1t, A₂, ..., A_k can be determined using the same way as case 1.

Recall, the integration of A/(ax + b)^rt, where r > 1, is

Therefore, the integration of the proper rational function is,

Case 3: Q(x) contains irreducible quadratic factors, none of which are repeated.

If Q(x) has a term ax² + bx + c, where b² - 4ac < 0, then in addition to case 1 and 2, the expression of P(x)/Q(x) will have a term of the form

      Ax+ B/(ax² + bx + c)

To integrate the above expression, (ax² + bx + c) needs to be rewritten as the following form by completing the square.

      ax² + bx + c = a(x+ d)² + e

and use the following formula

The next example shows how to integrate Ax+ B/(ax² + bx + c).

Make the substitution u = x -1, then du = dx, and x = (u + 1).

Case 4: Q(x) contains a repeated irreducible quadratic factor

If Q(x) has a factor (ax² + bx + c)^t, where b² - 4ac < 0, then the following terms will occur in the partial decomposition of P(x)/Q(x).

and each term in the above equation can be integrated by completing the square and making a tangent substitution, as shown in case 3.

The above methods work for proper rational functions. If the rational function P(x)/Q(x) is improper, which means the degree of P(x) is large than the degree of Q(x), the preliminary step of dividing Q(x) by P(x) until a remainder R(x) is obtained such that the degree of R(x) is smaller than Q(x). Then R(x)/Q(x) is a proper rational function and the above method can be used. The example on the left gives how to do the preliminary step of dividing.

Some functions can be changed into rational functions by substitutions and therefore integrated by the above method. When the integrand contains an expression of the form , the substitution may be effective. How to choose the substitution u depends on how the integrand looks. The following example gives the details of how to use the rationalizing substitutions.

In order to find the integration of , an appropriate substitution so that both roots in the denominator can be eliminated needs to be found. Since 6 is the least common multiple of 2 and 3, make the substitution , then dx = 6u⁵du.

This substitution changes the integrand to an improper rational function. The preliminary step of dividing (u + 1) by u³ gives

The function 1/(u +1) is a simple proper rational function with P(u) =1 and Q(u) = u+1.

where C is a constant.