Partial Fraction Expansion

The process of finding antiderivatives can feel a bit daunting as integrands become more elaborate. Luckily there are certain techniques that can help us identify and properly tackle more intricate indefinite integrals. In this topic, we'll take a look at the partial fraction expansion method, and how it can help us solve integrals with integrands comprised of rational functions.

Partial Fraction Expansion

This method works best for integrals of the form

$$\int \frac{P(x)}{Q(x)} \ dx$$where both $P$ and $Q$ are polynomials.

If the degree of $P$ is greater or equal than the degree of $Q$, we need to perform polynomial division so that

$$\frac{P(x)}{Q(x)} = C(x) + \frac{R(x)}{Q(x)}$$where $C$ and $R$ are also polynomials, and the degree of $R$ must be less than the degree of $Q$.

Once we have a proper rational expression where the degree of the numerator is less than the degree of the denominator, we can perform the partial fraction expansion as follows:

1. Factor the denominator into linear factors (and/or irreducible quadratic factors)
2. Make the rational expression equal to a sum of terms, each a partial fraction that corresponds to each factor in the denominator like so:

where $A, B, A_i, B_i$ are all constants that can be determined by comparison with the original expression. Afterwards, we are left with several much simpler integrals, usually with direct antiderivatives.

Factoring the denominator

Let's make this a bit less abstract with an example. Consider the following integral:

$$I = \int \frac{9x^2 + 3x - 2}{x^3 + 2x^2-2x+3} \ dx$$

We already have a proper rational expression here, since the degree of the denominator is 3 and the degree of the numerator is 2. So, we move on to factoring the denominator,

$$I = \int \frac{9x^2 + 3x - 2}{(x+3)(x^2 -x + 1)} \ dx$$

Obtaining the partial fraction expansion

Once the denominator is factored into linear factors and irreducible quadratic expressions, we can construct the integrand's partial fraction expansion as follows

$$\frac{9x^2 + 3x - 2}{(x+3)(x^2 -x + 1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2 -x + 1}$$In order to determine the constants $A, B, C$, let's first get rid of the denominators by multiplying both sides by it:

$$\begin{align*} 9x^2 + 3x - 2 &= \frac{A}{\cancel{x+3}} \cdot \cancel{(x+3)} \cdot(x^2 -x + 1) + \frac{Bx+C}{\cancel{x^2 -x + 1}} \cdot (x+3) \cdot (\cancel{x^2 -x + 1}) \\ 9x^2 + 3x - 2 &= A \cdot(x^2 -x + 1) + (Bx+C)\cdot(x+3) \\ 9x^2 + 3x - 2 &= Ax^2 -Ax + A + Bx^2+Cx + 3Bx+ 3C \end{align*}$$Now, we can compare the two expressions on both sides

$$9x^2 + 3x - 2 = (A+ B) \cdot x^2+ (-A + C + 3B) \cdot x + (A + 3C) \implies \begin{cases} 9 = A +B \\ 3 = -A + 3B +C \\ -2 = A + 3C \end{cases}$$After solving that system of equations, we are left with

$$A = \frac{70}{13} \qquad B = \frac{47}{13} \qquad C = -\frac{32}{13}$$

Finishing off using previous methods

Then,

$$\begin{align*} I = \int \frac{9x^2 + 3x - 2}{(x+3)(x^2 -x + 1)} \ dx &= \int \left( \frac{70/13}{x+3} + \frac{(47/13)x-(32/13)}{x^2 -x + 1} \right) \ dx \\ I &= \frac{70}{13} \int \frac{1}{x+3} \ dx + \frac{47}{13} \int \frac{x}{x^2 -x + 1} \ dx - \frac{32}{13} \int \frac{1}{x^2 -x + 1} \ dx \\ \end{align*}$$The integrals we are left with can be solved with both $u$-substitution (for the first two) and trigonometric substitution (for the last one):

For the first one, we have

$$I_1 = \frac{70}{13} \int \frac{1}{x+3} \ dx \implies I_1 = \frac{70}{13} \int \frac{1}{u} \ du = \frac{70}{13} \ln|u| + C \implies I_1 = \frac{70}{13} \ln|x+3| +C$$For remaining two, we will need to perform some algebraic manipulation first

$$\begin{align*} \frac{47}{2 \cdot 13} \int \frac{2x - 1 + 1}{x^2 -x + 1} \ dx - \frac{32}{13} \int \frac{1}{x^2 -x + 1} \ dx &= \frac{47}{26} \int \frac{2x - 1}{x^2 -x + 1} \ dx + \left( \frac{47}{26} - \frac{32}{13} \right) \int \frac{1}{x^2 -x + 1} \ dx \\ &= \frac{47}{26} \int \frac{2x - 1}{x^2 -x + 1} \ dx - \frac{17}{26} \int \frac{1}{x^2 -x + 1} \ dx = I_2 + I_3 \end{align*}$$So, for $I_2$ we have

$$I_2 = \frac{47}{26} \int \frac{2x - 1}{x^2 -x + 1} \ dx \implies I_2 = \frac{47}{26} \int \frac{1}{u} \ du = \frac{47}{26} \ln|u| + C \implies I_2 = \frac{47}{26} \ln|x^2 -x +1| +C$$And for $I_3$, we have

$$\begin{align*} I_3 = - \frac{17}{26} \int \frac{1}{x^2 -x + 1} \ dx = - \frac{17}{26} \int \frac{1}{\left(x-\frac{1}{2} \right)^2 + \frac{3}{4}} \ dx &= \left(- \frac{17}{26}\right)\left( \frac{4}{3} \right) \int \frac{1}{\left(\frac{2}{\sqrt{3}}x-\frac{1}{\sqrt{3}} \right)^2 + 1} \ dx \\ \implies I_3 &= - \frac{34}{39} \int \frac{1}{\tan^2(\theta) + 1} \cdot \left( \frac{\sqrt{3}}{2} \right) \cdot \sec^2(\theta) \ d\theta \\ &= \left( - \frac{34}{39} \right) \left( \frac{\sqrt{3}}{2} \right) \int \frac{1}{ \cancel{\sec^2(\theta)}} \cdot \cancel{\sec^2(\theta)} \ d\theta \\ &= - \frac{17\sqrt{3}}{39} \int d\theta = - \frac{17\sqrt{3}}{39} \theta + C \\ \implies I_3 &= - \frac{17\sqrt{3}}{39} \cdot \arctan\left(\frac{2}{\sqrt{3}}x-\frac{1}{\sqrt{3}} \right) + C \end{align*}$$Finally,

$$\int \frac{9x^2 + 3x - 2}{x^3 + 2x^2-2x+3} \ dx = \frac{70}{13} \ln|x+3| + \frac{47}{26} \ln|x^2 -x +1| - \frac{17\sqrt{3}}{39} \cdot \arctan\left(\frac{\sqrt{3}}{3} \cdot (2x- 1) \right) + C$$

Conclusion

In this topic, we have learned how to use the partial fraction expansion integration method. This method can be summarized as follows:

$$\int \frac{P(x)}{Q(x)} dx = \int \frac{A}{ax+b} dx + \int \frac{B} {cx+d} dx + \cdots$$where each summand in the right-hand side can be calculated using the methods mentioned in our previous topics. It should not come as a surprise to you that in order to master all these methods, you need a lot of practice. So, let's not waste time and proceed with tasks.