MATH 155 SUPPLEMENTAL NOTES 14

METHOD OF PARTIAL FRACTIONS

Suppose we would like to decompose the fraction 3/ 10. It is possible that this fraction was the result of the following addition problem.

If we can figure out the specific A and B that would satisfy this equation, then we could decompose this fraction.

The idea of decomposing a fraction is a useful technique in integration and this method is called the method of partial fractions. We will apply this method to rational expressions to expand the given expression as a sum of fractions with distinct factors for their denominators

 

Suppose that a factor of the denominator was the linear factor ax + b. What would be the derivative of this factor?

The derivative would be a constant.

So when we decompose the rational expression as a sum of fractions whose denominators are linear factors, the numerators will be constant.

Suppose that a factor of the denominator was the quadratic factor ax 2 + bx + c. What would be the derivative of this factor?

The derivative would be a linear factor.

So when we decompose the rational expression as a sum of fractions whose denominators might include an irreducible quadratic, the numerator will be a linear factor.

I will now break the types of fractions down into two types.

1.

Proper fractions - degree of the numerator < degree of the denominator

2.

Improper fractions - degree of the numerator ³ degree of the denominator

TYPE 1: PROPER FRACTIONS

I will now break this type down into three cases

CASE 1:

Distinct linear factors of multiplicity 1

CASE 2:

Linear factors of multiplicity of 2 or greater

CASE 3:

Irreducible quadratic factors

CASE 1: Distinct linear factors of multiplicity 1

I will step through the process as I solve the following example.

EXAMPLE 1:

SOLUTION:

Step 1:

Factor the denominator.

 

y 2 + y = y(y + 1)

Step 2:

Write the rational expression as a sum of the fractions whose denominators are the linear factors.

 

 

A and B are called undetermined coefficients.

Step 3:

Clear the fractions by multiplying both sides by the Least Common Denominator.

 

A(y + 1) + By = y + 4

or

Ay + A + By = y + 4

Step 4:

Collect all like terms together to form a system of equations.

 

Ay + By = y ® A + B = 1

A = 4

Step 5:

Solve for A and B.

 

A = 4 ® 4 + B = 1 ® B = -3

Therefore,

Step 6:

Substitute the sum in for the original integral and integrate.

 

 

Let u = y + 1, then du = dy.

 

EXAMPLE 2:

SOLUTION:

Step 1:

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

Step 2:

Step 3:

A(x + 2)(x - 1) + Bx(x - 1) + Cx(x + 2) = 1

or

Ax 2 + Ax - 2 + Bx 2 - Bx + Cx 2 + 2Cx = 1

Step 4:

Ax 2 + Bx 2 + Cx 2 = 0 ® A + B + C = 0

Ax - Bx + 2Cx = 0 ® A - B + 2C = 0

-2A = 1

Step 5:

 

 

Therefore,

Step 6:

CASE 2: Linear factors of multiplicity of 2 or greater

Here is how we will handle this case. Suppose we had a linear factor, ax + b, of multiplicity 2, the decomposition of the fraction would look like this:

If the linear factor were of multiplicity 3, the decomposition would look like this:

Notice that I started out with the linear factor raised to the first power, and then increased the power of the factor until I got to the multiplicity value. Also notice that each new denominator gets a different letter in the numerator.

EXAMPLE 3:

SOLUTION:

Step 1:

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

x - 1

Multiplicity 1

x + 1

Multiplicity 2

Step 2:

 

Noticed that I started with the non-repeated linear factor. This is not necessary, but it does help prevent confusion.

Step 3:

 

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

Ax 2 + 2Ax + A + Bx 2 - B + Cx - C = x 2

Step 4:

Ax 2 + Bx 2 = x 2 ® A + B = 1 ® B = 1 - A

2Ax + Cx = 0 ® 2A + C = 0 ® C = - 2A

A - B - C = 0

Step 5:

A - B + C = 0 ® A - (1 - A) - (-2A) = 0 ® 4A - 1 = 0 ® A = 1/4

 

Therefore,

 Step 6:

 

 

 Let u = x + 1, then du = dx.

 

 

 

 

EXAMPLE 4:

SOLUTION:

Step 1:

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

x

Multiplicity 1

x + 1

Multiplicity 2

Step 2:

Step 3:

 

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

Ax 2 + 2Ax + A + Bx 2 + Bx + Cx = x 2 + 1

Step 4:

Ax 2 + Bx 2 = x 2 ® A + B = 1

2Ax + Bx + Cx = 0 ® 2A + B + C = 0

A = 1

Step 5:

A = 1

1 + B = 1 ® B = 0

2 + 0 + C = 0 ® C = -2

Therefore,

Step 6:

 

Let u = x + 1, then du = dx.

 

CASE 3: Irreducible quadratic factors

What is an irreducible quadratic factor? It is a quadratic factor that cannot be factored into two linear factors. Now recall the discussion about the derivative of a quadratic factor. Remember it is a linear factor of the form ax + b. We will use this form for the numerator of the new fraction that has an irreducible quadratic factor in its denominator. Then we will proceed as before.

EXAMPLE 5:

SOLUTION:

Step 1:

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

Step 2:

 

Notice that x 2 + 1 is the irreducible quadratic factor, so the numerator will be the linear factor Bx + C. Also, notice that I put the linear factor first, this is not necessary, but it does keep things neat looking.

 Step 3:

 

 

 A(x 2 + 1) +x(Bx + C) = 1

Ax 2 + A + Bx 2 + Cx = 1

 Step 4:

Ax 2 + Bx 2 = 0 ®  A + B = 0

Cx = 0 ® C = 0

A = 1

 Step 5:

A = 1

C = 0

1 + B = 0 ® B = -1

Therefore,

 

 Step 6:

 

 

Let u = x 2 + 1, then du = 2x dx.

 

 

EXAMPLE 6:

SOLUTION:

Step 1:

x 4 + 5x 2 + 4 = (x 2 + 4)(x 2 + 1)

Step 2:

Step 3:

 

(Ax + B)(x 2 + 1) + (Cx + D)(x 2 + 4) = 4x 3 - 7x

Ax 3 + Ax + Bx 2 + B + Cx 3 + 4Cx + Dx 2 + 4D = 4x 3 - 7x

Step 4:

Ax 3 + Cx 3 = 4x 3 ® A + C = 4

Bx 2 + Dx 2 = 0 ® B + D = 0

Ax + 4Cx = -7x ® A + 4C = -7

B + 4D = 0

Step 5:

 

Therefore,

Step 6:

 

Let u = x 2 + 4, then du = 2dx, and let p = x 2 + 1, then dp = 2x dx.

 

TYPE 2: IMPROPER FRACTIONS

To handle this case we will have to use long division first to reduce the integrand into a polynomial and a proper fraction. After this is accomplished, then the resulting fractional integrand will fit into one of the three above cases or we will have to use a "trick" that we have already learned.

EXAMPLE 7:

SOLUTION:

First of all, we will perform long division to simplify this integrand.

Therefore,

and we will now proceed as if we were working on a proper fraction.

Step 1:

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

Step 2:

Step 3:

 

A(x - 1)(x + 1) + Bx(x + 1) +Cx(x - 1) = x + 1

Ax 2 - A + Bx 2 + Bx + Cx 2 - Cx = x + 1

Step 4:

Ax 2 + Bx 2 + Cx 2 = 0 ® A + B + C = 0

Bx - Cx = x ® B - C = 1

-A = 1

Step 5:

A = -1

-1 + B + C = 0 ® B + C = 1

1 + C = 1 ® C = 0

Therefore,

Step 6:

EXAMPLE 8:

SOLUTION:

Since it is an improper integral, I will first use long division to simplify the integrand.

Therefore,

Now to proceed as we did with proper fractions.

Step 1:

x 2 + 2x - 8 = (x + 4)(x - 2)

Step 2:

Step 3:

 

A(x - 2) + B(x + 4) = x

Ax - 2A + Bx + 4B = x

Step 4:

Ax + Bx = x ® A + B = 1

-2A + 4B = 0

Step 5:

Step 6:

 

I have provided you with a nice mixture of integrals that use the method of partial fractions. Work through these examples making note of how each case in type one is done. Then notice that the difference between how you would do type two from type one is that you have to perform long division first. Please feel free to contact me if you have any questions or problems with any of the examples in this set of supplemental notes.

RETURN TO INDEX