5.18 Integration Using Partial FractionsHL Only
A rational function is of the form $\dfrac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials. If the degree of the numerator $P(x)$ is strictly less than the degree of the denominator $Q(x)$, it is called a proper rational function. When dealing with complex proper rational functions, we can reverse the process of common denominators through a technique called partial fraction decomposition. This technique relies on factoring the denominator $Q(x)$ completely and writing the original fraction as a sum of simpler fractions, which can then be easily integrated using standard logarithmic, power, or inverse trigonometric rules.
Crucial Prerequisite: Partial fraction decomposition only works on proper rational functions. If the degree of the numerator is greater than or equal to the degree of the denominator, you must first use polynomial long division to express the integrand as a polynomial plus a proper rational function.
1. Distinct Linear Factors
When the denominator $Q(x)$ factors into distinct linear terms, such as $(x-a)(x-b)$, the decomposition takes the form:
$$\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$$
where $A$ and $B$ are unknown constants to be determined.
Example 1
Evaluate the integral $\displaystyle\int \dfrac{2}{x^2 - 1} dx$.
Solution:
First, factor the denominator: $x^2 - 1 = (x - 1)(x + 1)$. Next, set up the partial fraction decomposition $$\dfrac{2}{(x - 1)(x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}$$ Multiply through by the common denominator to equate the numerators $$2 = A(x + 1) + B(x - 1)$$ Let $x = 1$. Then, $$2 = 2A \implies A = 1$$ Let $x = -1$. Then, $$2 = -2B \implies B = -1$$ Now, substitute these constants back into the integral $$\begin{aligned} \int \dfrac{2}{x^2 - 1} dx &= \int \left( \dfrac{1}{x - 1} - \dfrac{1}{x + 1} \right) dx \\ &= \int \dfrac{1}{x - 1} dx - \int \dfrac{1}{x + 1} dx \\ &= \ln|x - 1| - \ln|x + 1| + c \\ &= \ln\left|\dfrac{x - 1}{x + 1}\right| + c \end{aligned}$$
2. Repeated Linear Factors
When the denominator contains a repeated linear factor, such as $(x-a)^n$, the decomposition must include a partial fraction for every power up to $n$:
$$\dfrac{P(x)}{(x-a)^n} = \dfrac{A_1}{x-a} + \dfrac{A_2}{(x-a)^2} + \dots + \dfrac{A_n}{(x-a)^n}$$
Example 2
Evaluate the integral $\displaystyle\int \dfrac{4x}{(x-1)^2(x+1)} dx$.
Solution:
Set up the partial fraction decomposition accounting for the repeated root $(x-1)$ and the distinct root $(x+1)$ $$\dfrac{4x}{(x-1)^2(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{C}{x+1}$$ Multiply through by the common denominator $(x-1)^2(x+1)$ to clear the fractions $$4x = A(x-1)(x+1) + B(x+1) + C(x-1)^2$$ We can find the constants by choosing strategic values for $x$. Let $x = 1$. Then, $$4(1) = 0 + B(2) + 0 \implies 4 = 2B \implies B = 2$$ Let $x = -1$. Then, $$4(-1) = 0 + 0 + C(-2)^2 \implies -4 = 4C \implies C = -1$$ To find $A$, substitute any other convenient value for $x$, such as $x = 0$, along with our known values for $B$ and $C$ $$0 = A(-1)(1) + 2(1) + (-1)(-1)^2 \implies 0 = -A + 2 - 1 \implies A = 1$$ Substitute $A=1, B=2, C=-1$ back into the integral and evaluate using logarithmic and power rules $$\begin{aligned} \int \dfrac{4x}{(x-1)^2(x+1)} dx &= \int \left( \dfrac{1}{x-1} + \dfrac{2}{(x-1)^2} - \dfrac{1}{x+1} \right) dx \\ &= \ln|x-1| + \int 2(x-1)^{-2} dx - \ln|x+1| \\ &= \ln|x-1| - \dfrac{2}{x-1} - \ln|x+1| + c \\ &= \ln\left|\dfrac{x-1}{x+1}\right| - \dfrac{2}{x-1} + c \end{aligned}$$
3. Irreducible Quadratic Factors
If the denominator contains a quadratic factor $ax^2+bx+c$ that cannot be factored further into real linear terms (its discriminant $b^2-4ac < 0$), the corresponding numerator in the decomposition must be a generic linear term $Ax+B$:
$$\dfrac{P(x)}{(x-r)(ax^2+bx+c)} = \dfrac{A}{x-r} + \dfrac{Bx+C}{ax^2+bx+c}$$
Integrating the resulting fraction $\dfrac{Bx+C}{ax^2+bx+c}$ often involves splitting it into two parts: one that evaluates to a natural logarithm via $u$-substitution, and one that evaluates to an arctangent function.
Example 3
Evaluate the integral $\displaystyle\int \dfrac{3x^2+x+4}{(x-1)(x^2+1)} dx$.
Solution:
The term $(x^2+1)$ is an irreducible quadratic. Set up the decomposition $$\dfrac{3x^2+x+4}{(x-1)(x^2+1)} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^2+1}$$ Multiply through by the common denominator $$3x^2+x+4 = A(x^2+1) + (Bx+C)(x-1)$$ Let $x = 1$. Then, $$3(1)^2+1+4 = A(1^2+1) + 0 \implies 8 = 2A \implies A = 4$$ Next, choose $x=0$ to quickly find $C$ $$4 = 4(1) + (0+C)(-1) \implies 4 = 4 - C \implies C = 0$$ Choose $x = -1$ to find $B$ $$3(-1)^2-1+4 = 4((-1)^2+1) + (-B+0)(-2) \implies B = -1$$ Substitute the constants into the integral. Note that because $C=0$, the second fraction is simplified, allowing for a straightforward $u$-substitution where $u = x^2+1$ and $du = 2x dx$ $$\begin{aligned} \int \dfrac{3x^2+x+4}{(x-1)(x^2+1)} dx &= \int \left( \dfrac{4}{x-1} - \dfrac{x}{x^2+1} \right) dx \\ &= 4\ln|x-1| - \dfrac{1}{2} \int \dfrac{2x}{x^2+1} dx \\ &= 4\ln|x-1| - \dfrac{1}{2}\ln(x^2+1) + c \end{aligned}$$
Notice that we wrote $\ln(x^2+1)$ without absolute value bars in the final line of Example 3. Since $x^2+1 > 0$ for all real numbers $x$, the absolute value operation is redundant and can be safely dropped.
4. Improper Rational Functions
Example 4
Evaluate the integral $\displaystyle\int \dfrac{x^3-x+3}{x^2+x-2} dx$.
Solution:
The degree of the numerator (3) is greater than the degree of the denominator (2), making this an improper rational function. We must first perform polynomial long division dividing $x^3-x+3$ by $x^2+x-2$ $$x^3-x+3 = x(x^2+x-2) - x^2 + x + 3$$ $$-x^2+x+3 = -1(x^2+x-2) + 2x + 1$$ Therefore, the division yields a quotient of $x-1$ and a remainder of $2x+1$. We can rewrite the integrand as $$\dfrac{x^3-x+3}{x^2+x-2} = x - 1 + \dfrac{2x+1}{x^2+x-2}$$ Now, apply partial fractions to the proper rational remainder $\dfrac{2x+1}{x^2+x-2}$. Factor the denominator: $x^2+x-2 = (x-1)(x+2)$ $$\dfrac{2x+1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$$ $$2x+1 = A(x+2) + B(x-1)$$ Let $x = 1$. Then, $$3 = 3A \implies A = 1$$ Let $x = -2$. Then, $$-3 = -3B \implies B = 1$$ Substitute this entire decomposition back into the original rewritten integral $$\begin{aligned} \int \dfrac{x^3-x+3}{x^2+x-2} dx &= \int \left( x - 1 + \dfrac{1}{x-1} + \dfrac{1}{x+2} \right) dx \\ &= \dfrac{x^2}{2} - x + \ln|x-1| + \ln|x+2| + c \\ &= \dfrac{x^2}{2} - x + \ln|(x-1)(x+2)| + c \end{aligned}$$