Page 35 - Calculus for the Clueless, Calc II

We now do the section I like the least. It is uninteresting, unimaginative, frequently overly long, and ...
necessary. Unless we have only linear factors, it is best to avoid this technique if possible.

We wish to do the integrals by partial fractions. Suppose we have R(x) = P(x)/Q(x), where the degree of P(x) is
less than the degree of Q(x). We wish to break up R(x) into simpler rational fractions; each piece is called a
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partial fraction. There will be one or more pieces for each linear factor x + a or quadratic factor x + b of Q(x).
Here's how it looks in a particular case:

Notice that each linear factor gives pieces with constants on top, and each quadratic factor gives pieces with
first-degree polynomials on top. The bottoms of the partial fractions are powers of the factors running from 1 to
the power that occurs in Q(x). The constants A, B, C, D, E, F, G, H, and I have to be solved for, which I hope
you never have to do. If you added all the fractions on the right, you would get the left fraction.

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One more thing. Suppose Ax + Bx + Cx + D = 4x - 7x - 1. Two polynomials are equal if their coefficients
match. So, A = 4, B = 0, C = -7, D = -1.
There are a number of techniques that will allow you to solve for A, B, C, and so on. Two of them
(combinations of) will serve us best.
Example 21—

Since the degree of the top is greater than or equal to the degree of the bottom, long-divide the
bottom into the top until the degree of the top is less than the degree of the bottom.

Look at the fractional part only.

We will solve for A and B in two different ways. We now add the fractions and equate the tops
since the bottoms are the same.

Method 1

Multiply out left side and group
terms.