Integration: Partial Fractions| Real Statistics Using Excel

We describe techniques to evaluate integrals of the form

where the numerator and denominator are polynomials. Our goal is to express the integrand as a sum of rational expressions, each of which can be integrated. For cases 1 through 4, below, we assume that the degree of u(x) is less than that of v(x).

Case 1: Denominator is a product of linear expressions

We assume that v(x) can be expressed as a product of linear factors v₁(x), …, v_k(x). We seek constants a₁, …, a_k such that the integrand can be expressed as a sum of expressions of the form

As a result, the integral can be evaluated as a sum of natural logs.

Example 1:

We seek A and B such that

Thus

A(x+3) + B(2x–1) ≡ 4x

Equivalently

(A+2B)x + (3A–B) ≡ 4x

It now follows that

A+2B = 4 3A–B = 0

Solving for A and B results in

A = 4/7 B = 12/7

Hence

We can also find the values for A and B by plugging x = –3 into the equation

A(x+3) + B(2x–1) ≡ 4x

to obtain –7B = –12, i.e. B = 12/7. Similarly, plugging in x = 1/2, we obtain 7A/2 = 2, i.e. A = 4/7.

Example 2:

This time

A+2B = 0 3A–B = 7

and so A = 2 and B = –1. Hence

Case 2: Denominator is a product of linear expressions with repetitions

If in case 1, one (or more) of the linear factors is repeated then the partial fractions used are slightly different. E.g. if v(x) = wⁿ(x) with n > 1, then

Actually, combinations of cases 1 and 2 are also supported.

Example 3:

We seek a solution such that

This means that

A(2x–1)(x+3) + B(x+3) + C(2x–1)² ≡ 2x² + 3

(2A+4C)x² + (5A+B–2C)x + (–3A+3B+C) = 2x² + 3

2A+4C = 2 5A+B–2C = 0 –3A+3B+C = 3

This results in A = 1/43, B = 37/43, and C = 21/43. Hence

Case 3: Denominator is a product of linear and quadratic expressions

If v(x) has a quadratic factor that can’t be factored into linear factors, then the approach is similar to case 1, except that the numerator for a quadratic factor is now linear and not a constant value.

For example, suppose v(x) can be expressed as a product of linear factors v₁(x), …, v_k(x) and one quadratic factor w(x). We seek constants a₁, …, a_k, b, c such that

Note that a quadratic of form ax² + bx + c can’t be factored if b² – 4ac < 0 (i.e. two imaginary roots).

Example 4:

First, we note that

x⁴ – 1 = (x² + 1)(x+1)(x – 1)

Thus, we seek A, B, and C such that

(Ax + B)(x + 1)(x – 1) + C(x² +1)(x – 1) + D(x² +1)(x + 1) ≡ x³ + 2

(A + C + D) x³ + (B – C + D) x² + (–A + C + D) x + (–B – C + D) ≡ x³ + 2

A + C + D = 1 B – C + D = 0 –A + C + D = 0 –B – C + D = 2

Solving these equations, we obtain A = 1/2, B = –1, C = –1/4, and D = 3/4. Hence

For the first integral, let u = x² + 1. Then du = 2x dx, and so

Thus

Case 4: Denominator factorization has a repeated quadratic factor

If in case 3, a quadratic term in v(x) is repeated n times, then the equivalence from case 3 becomes

Example 5:

(Ax + B)(x² + x +1) + (Cx + D) ≡ x³ – 2

A x³ + (A + B) x² + (A + B + C) x + (B + D) ≡ x³ – 2

A = 1 A + B = 0 A + B + C = 0 B + D = –2

Thus, A = 1, B = –1, C = 0, and D = –1. Hence

where c = 1/2 and a = √3/2. The first of these integrals can be evaluated as described in Example 11 of Quadratic Integrals. The second integral can be evaluated using the quadratic reduction formula (also described in Quadratic Integrals).

Case 5: Degree of u(x) ≥ v(x)

If the degree of u(x) is not less than the degree of v(x), then we need to employ long division to obtain an equivalent expression where the degree of the numerator is less than the degree of the denominator.