Lesson 0.1 - Review of Partial Fraction Decomposition

One of the biggest tools that I forgot how to use when going into my differential equations course was partial fraction decomposition.

Partial fraction decomposition is a method that we can use to take complex rational expressions and break them down into their simplest forms, which will be the sum or difference of multiple rational expressions. This tool helps us integrate these rational expressions.

We must recall one important integral in order to continue. In our explorations of single-variable integration calculus, we discovered that the following statement is true.

\int \frac{1}{x} d x = \ln | x |

We also discovered that this would be true for any constant $C$ in the numerator and a simple linear equation in the denominator.

Consider the following

\int \frac{2}{x + 3} = 2 \ln | x + 3 |

Now, what if we had something a bit more complicated. Let's take a look at a simple rational expression that does not allow us to use the previous method of integration.

Consider the expression

\frac{12}{x^{2} - 9}

Since the denominator of this equation is not a simple linear equation, we simply cannot integrate it the same way we had in the previous example. We must impose partial fraction decomposition.

In order to do so, we have a series of steps that we can follow.

Step 1: Completely factor the denominator into prime linear or quadratic factors. In our example above, the denominator is a difference of squares.

\frac{12}{(x - 3) (x + 3)}

Step 2: For linear factors: Each distinct linear factor of the form $(p x + q)$ include the following terms:

\frac{A}{(p x + q)}

For each repeated factor $(p x + q)^{n}$ , we must include n terms of the form:

\frac{A}{(p x + q)} + \frac{B}{(p x + q)} + \frac{C}{(p x + q)} + \dots + \frac{Z}{(p x + q)}

In this case, we have two linear terms in the denominator: $(x - 3)$ and $(x + 3)$ . Thus, we have

\frac{A}{x - 3} + \frac{B}{x + 3}

Step 3: For Quadratic Factors: Each distinct factor in the form $a x^{2} + b x + x$ include the following term:

\frac{B x + C}{(a x^{2} + b x + x)} + \frac{D x + E}{(a x^{2} + b x + x)} + \dots + \frac{Y x + Z}{(a x^{2} + b x + x)}

In the case above, we do not have quadratic factors in the denominator, so we can skip this step. However, we will go over an example of this later.

Step 4: Multiply each fraction by the LCD. Then solve the basic equation by determining the values of A,B,C,...,Z.

Before we get into step 4, let's look at what we've done so far.

\frac{12}{x^{2} - 9} = \frac{12}{(x - 3) (x + 3)} = \frac{A}{(x - 3)} + \frac{B}{(x + 3)}

Let's focus on the two right equations for step 4.

\frac{12}{(x - 3) (x + 3)} = \frac{A}{(x - 3)} + \frac{B}{(x + 3)}

The goal here is to cancel out our denominators so that we can get a linear equation to solve for A and B. To do so, let's multiply both sides by the lowest common denominator, which is $(x - 3) (x + 3)$ .

\frac{(x - 3) (x + 3)}{1} \times \frac{12}{(x - 3) (x + 3)} = (\frac{A}{(x - 3)} + \frac{B}{(x + 3)}) \times \frac{(x - 3) (x + 3)}{1}

Notice the denominators on both side cancel out, and we're left with

12 = A (x + 3) + B (x - 3)

Now that we have a simple linear expression, we can choose values of x that will cause A to become zero so we can solve for B, and vice versa. The clearest way to do this is to choose $x = 3$ and $x = - 3$ .

For $x = 3$

12 = A ((3) + 3) + B ((3) - 3)

For ( x = 3 ), we substitute 3 into the equation:

12 = A (3 + 3) + B (3 - 3)

This simplifies to:

12 = A (6) + B (0)

So, ( A = \frac{12}{6} = 2 ).

Now, for ( x = -3 ), substitute ( -3 ) into the equation:

12 = A ((- 3) + 3) + B ((- 3) - 3)

This simplifies to:

12 = A (0) + B (- 6)

So, $B = \frac{12}{- 6} = - 2$ .

Now that we know ( A = 2 ) and ( B = -2 ), we can write our partial fraction decomposition as:

\frac{12}{(x - 3) (x + 3)} = \frac{2}{x - 3} - \frac{2}{x + 3}

With this decomposition, we can now easily integrate the original expression. Applying the integral to both terms, we get:

\int \frac{12}{x^{2} - 9} d x = \int (\frac{2}{x - 3} - \frac{2}{x + 3}) d x

Using the integral formula we recalled earlier:

= 2 \ln | x - 3 | - 2 \ln | x + 3 | + C

Finally, we can combine these logarithmic terms if desired:

= 2 \ln (\frac{| x - 3 |}{| x + 3 |}) + C

The rational expression is decomposed, integrated, and simplified into its final form. This process of partial fraction decomposition can be applied to more complex rational expressions.

Example: Decomposing and Integrating with a Polynomial Numerator

Consider the expression:

\frac{2 x + 5}{x^{2} - x - 6}

Step 1: Factor the Denominator

First, factor the quadratic in the denominator:

x^{2} - x - 6 = (x - 3) (x + 2)

So, we rewrite the expression as:

\frac{2 x + 5}{(x - 3) (x + 2)}

Step 2: Set Up the Partial Fraction Decomposition

Since we have two distinct linear factors in the denominator, we set up the decomposition as:

\frac{2 x + 5}{(x - 3) (x + 2)} = \frac{A}{x - 3} + \frac{B}{x + 2}

Step 3: Multiply Both Sides by the LCD

Multiply both sides by the least common denominator, ( (x - 3)(x + 2) ), to eliminate the denominators:

2 x + 5 = A (x + 2) + B (x - 3)

Step 4: Solve for A and B

Expand the right-hand side:

2 x + 5 = A (x + 2) + B (x - 3)

2 x + 5 = A (x) + 2 A + B (x) - 3 B

Combine like terms:

2 x + 5 = (A + B) x + (2 A - 3 B)

Now, equate the coefficients of ( x ) and the constants from both sides. For the ( x )-terms:

A + B = 2

For the constant terms:

2 A - 3 B = 5

Step 5: Solve the System of Equations

We now solve the system of two equations:

( A + B = 2 )
( 2A - 3B = 5 )

From the first equation, solve for ( A ):

A = 2 - B

Substitute this into the second equation:

2 (2 - B) - 3 B = 5

Simplify:

4 - 2 B - 3 B = 5

4 - 5 B = 5

Solve for ( B ):

- 5 B = 1

B = - \frac{1}{5}

Substitute ( B = -\frac{1}{5} ) into ( A + B = 2 ):

A - \frac{1}{5} = 2

A = 2 + \frac{1}{5} = \frac{11}{5}

Step 6: Write the Decomposition

Now that we have ( A = \frac{11}{5} ) and ( B = -\frac{1}{5} ), we can write the partial fraction decomposition:

\frac{2 x + 5}{(x - 3) (x + 2)} = \frac{\frac{11}{5}}{x - 3} - \frac{\frac{1}{5}}{x + 2}

Or, equivalently:

\frac{2 x + 5}{(x - 3) (x + 2)} = \frac{11}{5 (x - 3)} - \frac{1}{5 (x + 2)}

Step 7: Integrate Both Terms

Now that the expression is decomposed, we can integrate it:

\int \frac{2 x + 5}{(x - 3) (x + 2)} d x = \int (\frac{11}{5 (x - 3)} - \frac{1}{5 (x + 2)}) d x

This integrates to:

= \frac{11}{5} \ln | x - 3 | - \frac{1}{5} \ln | x + 2 | + C

Finally, we can combine the logarithms if desired:

= \frac{1}{5} (11 \ln | x - 3 | - \ln | x + 2 |) + C

Example: Decomposing and Integrating with a Higher-Degree Polynomial Numerator

Consider the expression:

\frac{x^{2} + 5 x + 6}{(x - 1) (x^{2} + x + 1)}

Step 1: Factor the Denominator

The denominator already consists of one linear factor ( (x - 1) ) and one irreducible quadratic factor ( (x^2 + x + 1) ). Thus, no further factoring is necessary.

Step 2: Set Up the Partial Fraction Decomposition

Since we have one linear factor and one irreducible quadratic factor, we decompose the expression as follows:

\frac{x^{2} + 5 x + 6}{(x - 1) (x^{2} + x + 1)} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + x + 1}

Here, ( A ), ( B ), and ( C ) are constants that we need to determine.

Step 3: Multiply Both Sides by the LCD

Multiply both sides by the least common denominator, ( (x - 1)(x^2 + x + 1) ), to eliminate the denominators:

x^{2} + 5 x + 6 = A (x^{2} + x + 1) + (B x + C) (x - 1)

Step 4: Expand Both Sides

First, expand ( A(x^2 + x + 1) ):

A (x^{2} + x + 1) = A x^{2} + A x + A

Next, expand ( (Bx + C)(x - 1) ):

(B x + C) (x - 1) = B x (x - 1) + C (x - 1) = B x^{2} - B x + C x - C

So, we have:

x^{2} + 5 x + 6 = A x^{2} + A x + A + B x^{2} - B x + C x - C

Step 5: Combine Like Terms

Now, combine like terms on the right-hand side:

x^{2} + 5 x + 6 = (A + B) x^{2} + (A - B + C) x + (A - C)

Step 6: Solve for ( A ), ( B ), and ( C )

Now, equate the coefficients of ( x^2 ), ( x ), and the constant terms from both sides of the equation.

For the ( x^2 )-terms:

A + B = 1

For the ( x )-terms:

A - B + C = 5

For the constant terms:

A - C = 6

Step 7: Solve the System of Equations

We now solve the system of three equations:

( A + B = 1 )
( A - B + C = 5 )
( A - C = 6 )

From equation 3, solve for ( C ):

C = A - 6

Substitute this into equation 2:

A - B + (A - 6) = 5

Simplify:

2 A - B - 6 = 5

2 A - B = 11

Now, we have the system:

( A + B = 1 )
( 2A - B = 11 )

Add the two equations:

(A + B) + (2 A - B) = 1 + 11

3 A = 12

A = 4

Substitute ( A = 4 ) into ( A + B = 1 ):

4 + B = 1

B = - 3

Now, substitute ( A = 4 ) into ( C = A - 6 ):

C = 4 - 6 = - 2

Step 8: Write the Decomposition

Now that we have ( A = 4 ), ( B = -3 ), and ( C = -2 ), we can write the partial fraction decomposition:

\frac{x^{2} + 5 x + 6}{(x - 1) (x^{2} + x + 1)} = \frac{4}{x - 1} + \frac{- 3 x - 2}{x^{2} + x + 1}

Step 9: Integrate Both Terms

Now that the expression is decomposed, we can integrate it:

\int \frac{x^{2} + 5 x + 6}{(x - 1) (x^{2} + x + 1)} d x = \int \frac{4}{x - 1} d x + \int \frac{- 3 x - 2}{x^{2} + x + 1} d x

The first integral is straightforward:

\int \frac{4}{x - 1} d x = 4 \ln | x - 1 |

For the second integral, we use substitution. Let ( u = x^2 + x + 1 ), so ( du = (2x + 1) dx ). We rewrite the second integral as:

\int \frac{- 3 x - 2}{x^{2} + x + 1} d x

Step 10: Conclusion

After solving the integrals, the final result is:

\int \frac{x^{2} + 5 x + 6}{(x - 1) (x^{2} + x + 1)} d x = 4 \ln | x - 1 | + (solution for the second integral) + C

This shows how partial fraction decomposition works for a higher-degree polynomial numerator.

Next Lesson: Lesson 1 - What Are Differential Equations? What Should I Know Before Beginning?

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