Mathematics 136 – AP Calculus Integration by Partial Fractions Fall 2009

The partial fractions method applies to (in principle, all) rational functions

h(x) = \frac{f(x)}{g(x)} = \frac{a_{n}x^{n} + ⋯ + a_{1}x + a_{0}}{b_{m}x^{m} + ⋯ + b_{1}x + b_{0}}

(or to functions that can be brought to this form by a preliminary substitution). The steps involved are:

1. If n ≥ m, first divide g(x) into f(x) using polynomial division to write f(x) = q(x)g(x) + r(x), where the degree of r(x) is less than m. This yields

\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}

and

∫ \frac{f(x)}{g(x)} dx = ∫ q(x) dx + ∫ \frac{r(x)}{g(x)} dx

2. Now, assuming we have a rational function where the degree of the numerator is strictly less than the degree of the denominator, factor the denominator completely into linear and quadratic factors. The quadratic factors will have no real roots when they cannot be factored further. (It is a theorem that any polynomial with real coefficients can be factored in this way. The “bottleneck” in applying this method in very hard examples, though, is often finding the appropriate factorization. Needless to say, we will not be doing any of those very hard examples(!)

3. Set up the partial fractions.

a) If (x + a)^{e} is the highest power of a linear polynomial that divides g(x), then the partial fractions will include a group of terms

\frac{A_{1}}{x + a} + \frac{A_{2}}{(x + a)^{2}} + ⋯ + \frac{A_{e}}{(x + a)^{e}}

b) If (ax^{2} + bx + c) is a quadratic with no real roots that divides g(x), then the partial fractions will include a term

\frac{Bx + C}{ax^{2} + bx + c}.

Note the linear polynomial on the top. The reason you need both the Bx and the C will become apparent by doing some examples. It is not possible to “match” every possible numerator unless both the Bx and the C are included. Note: We will only be doing examples where quadratic factors appear to the first power in the factorization of the denominator. There is a more general form of partial fractions that applies if you have something like (x^{2}+6x+10)^{3} in the factorization, though. You need a sum of partial fractions in that case, something like in step 3 above.

4. Solve for the coefficients in the partial fractions. This can be done by clearing denominators, then substituting in x-values and/or equating coefficients on both sides of the resulting equation.

5. Integrate the partial fractions.

Worked Example

Here is an example illustrating many of these steps. Suppose we need to integrate

∫ \frac{x^{5} + 4x + 1}{x^{4} + 9x^{2}} dx.

The degree of the top is larger, so we divide first:

x^{5} + 4x + 1 = x(x^{4} + 9x^{2}) + (−9x^{3} + 4x + 1)

so

\frac{x^{5} + 4x + 1}{x^{4} + 9x^{2}} = x + \frac{−9x^{3} + 4x + 1}{x^{4} + 9x^{2}}

The denominator factors as x^{4} + 9x^{2} = x^{2}(x^{2} + 9). So the partial fractions are

\frac{−9x^{3} + 4x + 1}{x^{4} + 9x^{2}} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{Cx + D}{x^{2} + 9} .

Clearing denominators,

−9x^{3} + 4x + 1 = Ax(x^{2} + 9) + B(x^{2} + 9) + (Cx + D)x^{2}.

Substituting x = 0 we see 1 = 9B so B = 1/9. From the coefficient of x we see 4 = 9A, so A = 4/9. Then from the coefficients of x^{3}, A + C = −9, so C = −9 − 4/9 = −85/9 and finally from the coefficient of x^{2}, 0 = B + D, so D = −1/9. This gives

∫ \frac{x^{5} + 4x + 1}{x^{4} + 9x^{2}} dx = ∫ x + \frac{−9x^{3} + 4x + 1}{x^{4} + 9x^{2}} dx

= ∫ x + \frac{4/9}{x} + \frac{1/9}{x^{2}} + \frac{(−85/9)x + (−1/9)}{x^{2} + 9} dx

= \frac{x^{2}}{2} + \frac{4}{9} ln |x| − \frac{1}{9} \frac{1}{x} − \frac{85}{18} ln(x^{2} + 9) − \frac{1}{27} tan^{−1}(x/3) + C.

A final comment – note that if the Cx term were not present, then there would be no way to “match” the numerator −9x^{3} + 4x + 1 when we add up the partial fractions.