PARTIAL FRACTIONS: CASES AND EXAMPLES - MATHS - 2025

The partial fractions are fractions formed by polynomials, in which the denominator can be a linear or quadratic polynomial and, in addition, it can be raised to some power. Sometimes when we have rational functions it is very useful to rewrite this function as a sum of partial fractions or simple fractions.

This is so because in this way we can manipulate these functions in a better way, especially in cases where it is necessary to integrate said application. A rational function is simply the quotient between two polynomials, and they can be proper or improper.

If the degree of the polynomial of the numerator is less than the denominator, it is called a rational proper function; otherwise, it is known as an improper rational function.

Definition

When we have an improper rational function, we can divide the polynomial of the numerator by the polynomial of the denominator and thus rewrite the fraction p (x) / q (x), following the division algorithm as t (x) + s (x) / q (x), where t (x) is a polynomial and s (x) / q (x) is a proper rational function.

A partial fraction is any proper function of polynomials, whose denominator is of the form (ax + b) ⁿ or (ax ² + bx + c) ⁿ, if the polynomial ax ² + bx + c has no real roots and n is a number natural.

In order to rewrite a rational function in partial fractions, the first thing to do is factor the denominator q (x) as a product of linear and / or quadratic factors. Once this is done, the partial fractions are determined, which depend on the nature of these factors.

Cases

We consider several cases separately.

Case 1

The factors of q (x) are all linear and none are repeated. That is to say:

q (x) = (a ₁ x + b ₁) (a ₂ x + b ₂)… (a _s x + b _s)

There no linear factor is identical to another. When this case occurs we will write:

p (x) / q (x) = A ₁ / (a ₁ x + b ₁) + A ₂ / (a ₂ x + b ₂)… + A _s / (a _s x + b _s).

Where A ₁, A ₂,…, A _s are the constants to be found.

Example

We wish to decompose the rational function into simple fractions:

(x - 1) / (x ³ + 3x ² + 2x)

We proceed to factor the denominator, that is:

x ³ + 3x ² + 2x = x (x + 1) (x + 2)

Then:

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

(x - 1) / x (x + 1) (x + 2) = A / x + B / (x + 1) + C / (x + 2)

Applying least common multiple, it can be obtained that:

x - 1 = A (x + 1) (x + 2) + B (x + 2) x + C (x + 1) x.

We want to obtain the values ​​of the constants A, B and C, which can be found by substituting the roots that cancel each of the terms. Substituting 0 for x we ​​have:

0 - 1 = A (0 + 1) (0 + 2) + B (0 + 2) 0 + C (0 + 1) 0.

- 1 = 2A

A = - 1/2.

Substituting - 1 for x we ​​have:

- 1 - 1 = A (- 1 + 1) (- 1 + 2) + B (- 1 + 2) (- 1) + C (- 1 + 1) (- 1).

- 2 = - B

B = 2.

Substituting - 2 for x we ​​have:

- 2 - 1 = A (- 2 + 1) (- 2 + 2) + B (- 2 + 2) (- 2) + C (- 2 + 1) (- 2).

–3 = 2C

C = –3/2.

In this way the values ​​A = –1/2, B = 2 and C = –3/2 are obtained.

There is another method to obtain the values ​​of A, B and C. If on the right side of the equation x - 1 = A (x + 1) (x + 2) + B (x + 2) x + C (x + 1) x we ​​combine terms, we have:

x - 1 = (A + B + C) x ² + (3A + 2B + C) x + 2A.

Since this is an equality of polynomials, we have that the coefficients on the left side must be equal to those on the right side. This results in the following system of equations:

A + B + C = 0

3A + 2B + C = 1

2A = - 1

Solving this system of equations, we obtain the results A = –1/2, B = 2, and C = -3/2.

Finally, substituting the values ​​obtained we have that:

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

Case 2

The factors of q (x) are all linear and some are repeated. Suppose that (ax + b) is a factor that repeats "s" times; then, to this factor correspond the sum of «s» partial fractions.

A _s / (ax + b) ^s + A _s-1 / (ax + b) ^s-1 +… + A ₁ / (ax + b).

Where A _s, A _s-1,…, A ₁ are the constants to be determined. With the following example we will show how to determine these constants.

Example

Decompose into partial fractions:

(x - 1) / (x ² (x - 2) ³)

We write the rational function as a sum of partial fractions as follows:

(x - 1) / (x ² (x - 2) ³) = A / x ² + B / x + C / (x - 2) ³ + D / (x - 2) ² + E / (x - 2).

Then:

x - 1 = A (x - 2) ³ + B (x - 2) ³ x + Cx ² + D (x - 2) x ² + E (x - 2) ² x ²

Substituting 2 for x, we have that:

7 = 4C, that is, C = 7/4.

Substituting 0 for x we ​​have:

- 1 = –8A or A = 1/8.

Substituting these values ​​in the previous equation and developing, we have that:

x - 1 = 1/8 (x ³ - 6x ² + 12x - 8) + Bx (x ³ - 6x ² + 12x - 8) + 7 / 4x ² + Dx ³ - 2Dx ² + Ex ² (x ² - 4x + 4)

x - 1 = (B + E) x ⁴ + (1/8 - 6B + D - 4E) x ³ + (- ¾ + 12B + 7/4 - 2D + 4E) x ² + (3/2 - 8B) x - 1.

Equating coefficients, we obtain the following system of equations:

B + E = 0;

1/8-6B + D-4E = 1;

- 3/4 + 12B + 7/4 - 2D + 4E = 0

3/2 - 8B = 0.

Solving the system, we have:

B = 3/16; D = 5/4; E = - 3/16.

For this, we have to:

(x - 1) / (x ² (x - 2) ³) = (1/8) / x ² + (3/16) / x + (7/4) / (x - 2) ³ + (5 / 4) / (x - 2) ² - (3/16) / (x - 2).

Case 3

The factors of q (x) are linear quadratic, without any repeated quadratic factors. In this case, the quadratic factor (ax ² + bx + c) will correspond to the partial fraction (Ax + B) / (ax ² + bx + c), where the constants A and B are the ones to be determined.

The following example shows how to proceed in this case

Example

Decompose into simple fractions a (x + 1) / (x ³ - 1).

First we proceed to factor the denominator, which gives us as a result:

(x - 1) = (x - 1) (x + x +1).

We can observe that (x ² + x + 1) is an irreducible quadratic polynomial; that is, it does not have real roots. Its decomposition into partial fractions will be as follows:

(x + 1) / (x - 1) (x ² + x +1) = A / (x - 1) + (Bx + C) / (x ² + x +1)

From this we obtain the following equation:

x + 1 = (A + B) x ² + (A - B + C) x + (A - C)

Using equality of polynomials, we obtain the following system:

A + B = 0;

A-B + C = 1;

A-C = 1;

From this system we have that A = 2/3, B = - 2/3 and C = 1/3. Substituting, we have that:

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

Case 4

Finally, case 4 is the one in which the factors of q (x) are linear and quadratic, where some of the linear quadratic factors are repeated.

In this case, if (ax ² + bx + c) is a quadratic factor that repeats "s" times, then the partial fraction corresponding to the factor (ax ² + bx + c) will be:

(A ₁ x + B) / (ax ² + bx + c) +… + (A _s-1 x + B _s-1) / (ax ² + bx + c) ^s-1 + (A _s x + B _s) / (ax ² + bx + c) ^s

Where the A _s, A _s-1,…, A and B _s, B _s-1,…, B are the constants to be determined.

Example

We want to decompose the following rational function into partial fractions:

(x - 2) / (x (x ² - 4x + 5) ²)

Since x ² - 4x + 5 is an irreducible quadratic factor, we have that its decomposition into partial fractions is given by:

(x - 2) / (x (x ² - 4x + 5) ²) = A / x + (Bx + C) / (x ² - 4x +5) + (Dx + E) / (x ² - 4x + 5) ²

Simplifying and developing, we have:

x - 2 = A (x ² - 4x + 5) ² + (Bx + C) (x ² - 4x + 5) x + (Dx + E) x

x - 2 = (A + B) x ⁴ + (- 8A - 4B + C) x ³ + (26A + 5B - 4C + D) x ² + (- 40A + 5C + E) x + 25A.

From the above we have the following system of equations:

A + B = 0;

- 8A - 4B + C = 0;

26A + 5B - 4C + D = 0;

- 40A + 5C + E = 1;

25A = 2.

When solving the system, we are left with:

A = - 2/25, B = 2/25, C = - 8/25, D = 2/5 and E = - 3/5.

By substituting the obtained values ​​we have:

(x - 2) / (x (x ² - 4x + 5) ²) = -2 / 25x + (2x - 8) / 25 (x ² - 4x +5) + (2x - 3) / 5 (x ² - 4x + 5) ²

Applications

Integral calculus

Partial fractions are used primarily for the study of integral calculus. Here are some examples of how to perform integrals using partial fractions.

Example 1

We wish to calculate the integral of:

We can see that the denominator q (x) = (t + 2) ² (t + 1) is made up of linear factors where one of these is repeated; This is why we are in case 2.

We have to:

1 / (t + 2) ² (t + 1) = A / (t + 2) ² + B / (t + 2) + C / (t + 1)

We rewrite the equation and we have:

1 = A (t + 1) + B (t + 2) (t + 1) + C (t + 2) ²

If t = - 1, we have:

1 = A (0) + B (1) (0) + C (1)

1 = C

If t = - 2, it gives us:

1 = A (- 1) + B (0) (- 1) + C (0)

A = - 1

Then, if t = 0:

1 = A (1) + B (2) (1) + C (2)

Substituting the values ​​of A and C:

1 = - 1 + 2B + 4

1 = 3 + 2B

2B = - 2

From the above we have that B = - 1.

We rewrite the integral as:

We proceed to solve it by the substitution method:

This is the result:

Example 2

Solve the following integral:

In this case we can factor aq (x) = x ² - 4 as q (x) = (x - 2) (x + 2). We are clearly in case 1. Therefore:

(5x - 2) / (x - 2) (x + 2) = A / (x - 2) + B / (x + 2)

It can also be expressed as:

5x - 2 = A (x + 2) + B (x - 2)

If x = - 2, we have:

- 12 = A (0) + B (- 4)

B = 3

And if x = 2:

8 = A (4) + B (0)

A = 2

Thus, we are left with solving the given integral is equivalent to solving:

This gives us as a result:

Example 3

Solve the integral:

We have q (x) = 9x ⁴ + x ², which we can factor into q (x) = x ² (9x ² + 1).

This time we have a repeated linear factor and a quadratic factor; that is, we are in case 3.

We have to:

1 / x ² (9x ² + 1) = A / x ² + B / x + (Cx + D) / (9x ² + 1)

1 = A (9x ² + 1) + Bx (9x ² + 1) + Cx ² + Dx ²

Grouping and using equal polynomials, we have:

1 = (9B + C) x + (9A + D) x + Bx + A

A = 1;

B = 0;

9A + D = 0;

9B + C = 0

From this system of equations we have:

D = - 9 and C = 0

In this way, we have:

By solving the above, we have:

Law of mass action

An interesting application of the partial fractions applied to the integral calculus is found in chemistry, more precisely in the law of mass action.

Suppose we have two substances, A and B, which join together and form a substance C, so that the derivative of the amount of C with respect to time is proportional to the product of the amounts of A and B at any given moment.

We can express the law of mass action as follows:

In this expression α is the initial number of grams corresponding to A and β the initial number of grams corresponding to B.

Furthermore, r and s represent the number of grams of A and B respectively that combine to form r + s grams of C. For its part, x represents the number of grams of substance C at time t, and K is the constant of proportionality. The above equation can be rewritten as:

Making the following change:

We have that the equation becomes:

From this expression we can obtain:

Where if a ≠ b, partial fractions can be used for integration.

Example

Let's take for example a substance C that arises from combining a substance A with a B, in such a way that the mass law is fulfilled where the values ​​of a and b are 8 and 6 respectively. Give an equation that gives us the value of grams of C as a function of time.

Substituting the values ​​in the given mass law, we have:

When separating variables we have:

Here 1 / (8 - x) (6 - x) can be written as the sum of partial fractions, as follows:

Thus, 1 = A (6 - x) + B (8 - x)

If we substitute 6 for x, we have B = 1/2; and substituting 8 for x, we have A = - 1/2.

Integrating by partial fractions we have:

This gives us as a result:

Differential equations: logistic equation

Another application that can be given to partial fractions is in the logistic differential equation. In simple models we have that the growth rate of a population is proportional to its size; that is to say:

This case is an ideal and is considered realistic until it happens that the resources available in a system are insufficient to support the population.

In these situations, the most reasonable thing is to think that there is a maximum capacity, which we will call L, that the system can sustain, and that the growth rate is proportional to the size of the population multiplied by the available size. This argument leads to the following differential equation:

This expression is called the logistic differential equation. It is a separable differential equation that can be solved with the partial fraction integration method.

Example

An example would be to consider a population that grows according to the following logistic differential equation y '= 0.0004y (1000 - y), whose initial data is 400. We want to know the size of the population at time t = 2, where t is measured in years.

If we write y 'with Leibniz's notation as a function that depends on t, we have:

The integral on the left side can be solved using the partial fraction integration method:

We can rewrite this last equality as follows:

- Substituting y = 0 we have that A is equal to 1/1000.

- Substituting y = 1000 we have that B is equal to 1/1000.

With these values ​​the integral is as follows:

The solution is:

Using the initial data:

When clearing and we have:

Then we have that at t = 2:

In conclusion, after 2 years the population size is approximately 597.37.

References

1. A, RA (2012). Mathematics 1. Universidad de los Andes. Publications Council.
2. Cortez, I., & Sanchez, C. (nd). 801 Resolved integrals. National Experimental University of Tachira.
3. Leithold, L. (1992). The calculation with analytic geometry. HARLA, SA
4. Purcell, EJ, Varberg, D., & Rigdon, SE (2007). Calculation. Mexico: Pearson Education.
5. Saenz, J. (nd). Integral calculus. Hypotenuse.