Mathematics 132 – Calculus for Physical and Life Sciences 2 Integration by Partial Fractions February 11 - 12, 2008

The partial fraction method applies to 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.

3. Set up the partial fractions. 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}}

If (ax^{2} + bx + c)^{f} is the highest power of a quadratic with no real roots that divides g(x), then the partial fractions will include a group of terms

\frac{B_{1}x + C_{1}}{ax^{2} + bx + c} + \frac{B_{2}x + C_{2}}{(ax^{2} + bx + c)^{2}} + ⋯ + \frac{B_{f}x + C_{f}}{(ax^{2} + bx + c)^{f}} .

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.

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.