In this topic you are going learn how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called **partial fractions**, that is simplified and easy for us to integrate. The process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called **partial fraction decomposition**. It is important algebraic technique because many integrals involving rational expressions can be done if we first do partial fractions on the integrand.

This method is based on the simple concept of adding fractions by getting a common denominator.

The Partial Fractions decomposition for ¾ is

This concept can also be used with functions of . For example,

so that we can now say that a partial fractions decomposition for is

Integrating the function:

### METHOD: In general Consider a rational function:

where **P **and **Q **are polynomials. It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q: Such a rational function is called proper. If f is improper, then we must take the preliminary step of dividing Q into P (by long division) until a remainder R(x) is obtained such that deg(R) < deg(Q): The division statement is

where **S **and **R **are also polynomials. Sometimes this **preliminary step** is all that is required to get the integral.

Example 1. Evaluate

SOLUTION: Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write

**The next step** is to factor the denominator Q(x) as far as possible. It can be shown that any polynomial Q can be factored as a product of linear factors (of the form ax + b) and irreducible quadratic factors (of the form ax2 + bx + c; where b2 – 4ac < 0). For instance,

- x2 – 1 = (x – 1) (x + 1)
- 10×2 – 11x – 6 = (2x – 3) (5x + 2)
- x3 + 1 = (x + 1) (x2 – x + 1)
- x3 + 5 = (x + 5⅓) (x2 – 5⅓x + 25⅓)
- x3 + x2 + x + 1 = x3 (x + 1) + x + 1 = (x + 1) (x3 + 1)
- 3×3 + 14×2 + 7x – 4 = 3×2 (x + 1) + 11x(x + 1) – 4 (x + 1) = (x + 1) (3x – 1) (x + 4)
- x4 – 16 = (x2)2 – 42 = (x2 – 4) (x2 + 4) = (x – 2) (x + 2) (x2 + 4)
- x4 + 16 = x4 + 8×2 + 16 – 8×2 = (x2 + 4)2 – 8×2 = (x2 + 2√2x + 4) (x2 – 2 √2x + 4)
- x5 + 1 = (x + 1)(x4 – x3 + x2 – x + 1) = ¼ (x + 1) (2×2 – x + √5x + 2) (2×2 – x – √5x + 2)
- x6 – 3×5 + 10×3 – 15×2 + 9x – 2 = (x + 2) (x – 1)5
- x9 + 6×8 + 21×7 + 51×6 + 81×5 + 87×4 + 32×3 – 63×2 – 108x – 108 = (x – 1) (x + 2)2 (x2 + x + 3)3
- x2 + 1; x2 + 2x + 5; etc. .. are irreducible quadratic factors

**The third step** is to express the proper rational function R(x)/Q(x) as a sum of partial fractions of the form

A theorem in algebra guarantees that it is always possible to do this. There are **four possible cases**:

### Case I: The denominator Q(x) is a product of distinct linear factors.

This means that we can write

where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants A1;A2; . . . ;Ak such that

Example 2: Consider

**SOLUTION:** Since the degree of the numerator is less than the degree of the denominator,

we don’t need to divide. We factor the denominator as

**2×3 + 3×2 – 2x = x( 2×3 + 3x – 2) = x ( 2x – 1 ) (x + 2 )**

As you can see, the denominator has three distinct linear factors (with yellow background), therefore, the partial fraction decomposition of the integrand has the form

To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators,**x ( 2x – 1 ) (x + 2 )** , obtaining

**x2 + 2x – 1 = A (2x – 1) ( x + 2) + Bx ( x + 2) + Cx ( 2x – 1)**

Expanding the right side of Equation and writing it in the standard form for polynomials, we get

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

The polynomials in Equation are identical, so their coefficients must be equal. The coefficient of **x2** on the right side, **2A + B + 2C**, must be equal to the coefficient of **x****2** on the left side—which is, **1** . Likewise, the coefficients of **x** are equal and the constant terms are equal. This gives the following system of equations for A, B, and C:

**2A + B + 2C = 1**

**3A + 2B – C = 2**

**-2A = -1**

Solving, we get A = ½ , B = ⅕ , C = -1/10, integrate the partial fractions

At this point, it is a must that you can already do simple integration in mind. In integrating the middle term you can made the mental substitution * u* = 2x – 1, which gives

*= 2 dx and dx =*

**du****½**

*du***Note**: Another way of finding the coefficients of A, B and C is elimination or simplifying the equation

**x2 + 2x – 1 = A (2x – 1) ( x + 2) + Bx ( x + 2) + Cx ( 2x – 1)**

Let x = 0, the second and third term eliminate, the equation becomes -1 = -2A, therefore **A = 1/2**

Let x = 1/2, the first and third terms eliminate, 1/4 = 5/4B, then **B = 1/5**

Let x = -2, the first and second terms eliminate, -1 = 10C, we get **C = -1/10**

**Important:** After getting the coefficients of A, B, and C, substitute then proceed integrating the partial fractions. As you can see this method of getting the coefficients is much simpler and easy compare to the previous one but it is up to you what do you prefer.

### Case II: The denominator Q(x) is a product of linear factors, some of which are repeated.

This means that we can write

Instead of the single term A1/(a1x + b1x) in (Case I), we would use

Example 3: Find

Solution: The Partial Fraction decomposition is

Putting the right side over the common denominator (x – 1)3, you have

Equating numerators, (I’m going to use the second method then the first method.)

**3×2 – x + 4 = A(x – 1)2 + B(x – 1) + C**

Let x = 1, first and second terms eliminate, equation equal to, 3 -1 + 4 = C, thus, **C = 6**

Equating coefficients of **x2 **: **3 = A**

Equating coefficients of **x**: -1 = -2A + B ; B = -1 + 2(3) ; thus **B = 5**

Evaluate:

### Case III: The denominator Q(x) contains irreducible quadratic factors, none of which is repeated.

For example: x2 + 1; x(2x – 3) (x2 + x + 1); x2(x2 + 4) (x2 + 3x + 8); etc.

In this case if Q(x) has the factor ax2 +bx + c; where b2 – 4ac < 0; then, in addition to the partial fractions in (Case I) and (Case II), the expression for R(x)/Q(x) will have a term of the form

**A**and

**B**are constants to be determined.

Example 4: Evaluate

**SOLUTION:** Factored the denominator, x3 + 4x = x (x2 + 4 ) thus, the partial fraction decomposition is

Multiplying by x (x2 + 4), we have

**2×2 – x + 4 = A (x2 + 4) + (Bx + C) x**

**= (A + B)**

**x2 + Cx + 4A**

Equating these coefficients

Coefficients of x2 : 2 = (A + B)

Coefficients of x : -1 = C

Coefficients of x0 : 4 = 4A

Then, from 3rd equation **A = 1** , from 1st equation using A = 1 we got **B = 1** , and from the second equation we have **C = -1**

Hence,

=

**Important:** To integrate the second term I split it into two parts. Make the u-substitution for the second integral so that ** du = 2xdx. **Evaluate the third integral using the Inverse Trigonometric forms of tangent. Trigonometric Table/Formulas.

### Case IV: The denominator Q(x) contains a repeated irreducible quadratic factor.

For example: (x2 + x + 2)2; (2x + 7)(x2 + 4)3; x(x – 2)(x + 1)2(2×2 + 3)(x2 + 1)3(x2 + 4x + 5)4; etc.

In this case if Q(x) has the factor (ax2 + bx + c)r; where b2 – 4ac < 0; then, instead of a single partial fraction (In Case III), the sum

occurs in the partial fraction decomposition of R(x)/Q(x). Each of the terms in the decomposition can be integrated by using a substitution or by first completing the square if necessary.

Example 5:

**SOLUTION**: The form of the partial fraction decomposition is

Then,

**-x3 + 2×2 – x + 1 = A (x2 + 1)2 + ( Bx + C ) x ( x2 + 1) + ( Dx + E ) x**

**= A( x4 + 2×2 + 1) + (B x4 + x2) + C( x3 + x ) + Dx2 + Ex**

**= ( A + B) x4 + Cx3 + ( 2A + B + D )x2 + ( C + E )x + A**

Equating the coefficients,

Coefficients of x4: 0 = A + B

Coefficients of x3: C = -1

Coefficients of x2: 2 = 2A + B + D

Coefficients of x1: -1 = C + E

Coefficients of x0: 1 = A

Then solving we get, A = 1 , B = -1 , C = -1 , D = 1 , and E = 0

Evaluate:

=

**Important**:Again, you will notice that I split the second integral. Try some practice exercises to familiarize yourself to evaluate integral using this method. Integration by Partial Fractions – Set 1 Problems

*credit: D. A. Kouba (UC Davis), Kiryl Tsishchanka, James Stewart**©2013 www.PinoyBIX.com*

If you liked this, then please subscribe to our YouTube Channel for engineering video tutorials. You can also find us on Twitter and Facebook.