Partial Fractions Day 2 Chapter 7.4 April 3, 2006

Integrate:

The Arctangent formula (also see day 9 notes) The “new” formula: When to use? If the polynomial in the denominator does not have real roots (b 2 -4ac < 0) then the integral is an arctangent, we complete the square and integrate…. For example:

What if the polynomial has real roots? That means we can factor the polynomial and “undo” the addition! To add fractions we find a common denominator and add: we’ll work the other way…. The denominator of our rational function factors into (t + 4)(t -1) So in our original “addition,” the fractions were of the form:

What if the polynomial has real roots? We now need to solve for A and B. Multiplying both sides by the denominator: Choose t to be 1 and substituting into the equation above, we get 1 = 5B or B = 1/5 And with t = -4: 1 = -5A, or A = -1/5

Integrating with the Partial Fraction Decomposition: Our Rational function can be written as: And the integral becomes: Which integrates to:

Examples:

Each of these integrals involved linear factors What if a factor is repeated? For example: The “x” factor is repeated, so in our original addition, we could have had each of the “reduced” fractions: Clearing our denominators, we get:

To Solve for A, B, and C, again we choose x carefully: If we let x = 0, we can solve for B, Next, if x=-1, we can solve for C, We still need “A.” But because we know B and C, we can choose any x and use the known values to solve for A. Let x = 1

Using this information, our original integral becomes: We can integrate each term resulting in: Another form of the answer might be:

Example: Undoing the addition, we get: Solving for A, B and C, we arrive at: Integrating, we get:

We may also have expressions with factors of higher powers: We apply the same concept as when there are linear factors, we undo the addition using REDUCED fractions. Clearing the denominator, we get: Letting x = 0, we can solve for A: Solving for B and C, we need a new strategy……… The highest power on top must be less than the power on the bottom.

We have To solve for B and C, we will match coefficients From Matching the coefficients of both sides of the equation, term by term, we get: Returning to our original integral, we get that: Integrating the -2/x is clear, but what about the second term?

The Integration: We also will need to split the remaining integral. The first integral can be solved using u-substitution Splitting the integral, we get:

The Integration: The first integral can be solved using u-substitution The second integral is an arctangent Final Solution:

More Examples:

In groups, try: