PARTIAL FRACTIONS (BC TOPIC ONLY) - Advanced Methods of Integration - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC

Master AP Calculus AB & BC


CHAPTER 8. Advanced Methods of Integration


Integration by partial fractions is a method used to simplify integrals based on a very cool trick. The trick is so unique that it tends to stick with you, so you shouldn’t have any trouble remembering how it’s done come test time. That’s good news, especially since the test is such a high-pressure situation that most people forget important things, such as what their name is, when the Magna Carta was signed, what the current exchange rate is between major world currencies, and what exactly that little symbol is that became The Artist Formerly Known as Prince’s new name.

Partial fraction decomposition allows you to break a rational (fractional) expression into the sum of a couple of smaller fractions. Here’s the great thing: The denominators of the smaller fractions are the factors of the original denominator. The numerators of those smaller fractions are just constants. It’s your job to find out what they are exactly.

Example 12: Evaluate

Solution: You may be tempted to try a u-substitution with u = 2x2 — 9x — 5 and to try and force the numerator into du = (4x — 9)dx (as we did earlier in this chapter). However, you should use partial fractions because the denominator is factorable.

Partial fraction decomposition tells us that

In other words, we can find two constants A and B such that the sum of the two right fractions equals the larger fraction on the left. How do you do that? First, eliminate the fractions by multiplying both sides of the above equation by (2x + 1)(x — 5).

x + 3 = A(x — 5) + B(2x + 1)

Now, distribute the constants.

x + 3 = Ax — 5A + 2Bx + B

You’re almost finished; factor the x out of the Ax and 2Bx terms.

x + 3 = (A + 2B)x — 5A + B

Stop and look at that for a second. If the two sides are equal, then A + 2B (the coefficient of the x on the right side) must be equal to 1 (the coefficient of the x on the left side). Similarly, —5A + B must equal 3. Therefore, you get the system of equations:

A + 2B = 1

—5A + B = 3

NOTE. When you integrate by partial fractions on the AP test, the denominators will always have linear factors. This technique is slightly modified when the factors have higher degrees, but you don’t have to worry about that for the AP test.

Use whatever technique you want to simultaneously solve these equations (linear combination, substitution, matrices, etc.) to figure out that A = —5/11 and B = 8/11. Therefore,

and instead of integrating the ugly left side, we can integrate the slightly less ugly right side.

Use u-substitution in each integral with u = the denominator of each and the result is

or if you feel like getting common denominators and going nuts with log properties, you can rewrite as

TIP. Even though the A and B ended up looking kind of gross, integrating was still quite easy. That’s why integrating by partial fractions is so great.

I have no idea why you would ever want to do that, but hey, whatever floats your boat. In conclusion, you should integrate by partial fractions if you can factor the denominator. Create a sum of new fractions such that the denominators of the new fractions are the factors of the original denominator and the numerators of the new fractions are constants. Once you determine what those constants must be, all that remains is to integrate the string of smaller fractions, which is typically very easy.


Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.



1. The denominator factors to (x + 4)(x + 5); to begin partial fractions, you set up the following equation:

Multiply through by (x + 4)(x + 5) to eliminate fractions, and then find A and B.

x — 3 = A(x + 5) + B(x + 4)

x — 3 = (A + B)x + 5A + 4B

This results in the system of equations:

A + B — 1 and 5A + 4B = —3

A = —7, B = 8

The original integral now becomes

2. This problem does not require partial fractions. If you factor the denominator, the fraction simplifies.

This is an easy u-substitution problem if u = x — 4.

ln |x — 4| + C

3. You can pull the 4 out of the integral if you want (and replace it with a 1), but that doesn’t make the problem significantly easier. Because there is no x in the numerator, its coefficient must be 0 (that’ll be important in a few seconds). Factoring the denominator may be the hardest part of this problem.

This leads to the system A + 3B = 0 and —2A + 7B = 4, whose solution is A = —12/13 and B = 4/13.

4. The denominator has three factors this time, but the technique stays exactly the same.

The only constant term on the right is —6A, so —6A = 1 and A = —1/6. There is no squared term on the left, so (after substituting in the value of A) you have the system

After solving the system, you get A = —1/6, B = —1/3 and C = 1/2, making the integral