Partial Fractions – Definition, Condition, and Examples

Integration of rational functions using partial functions is an essential technique to add to your integration tool kits. This method comes in handy when the substitution method can’t be used to integrate a specific rational function.

Partial fraction decomposition allows us to integrate complex rational functions. The integration technique shows us how to break down rational functions as sums of simpler functions then integrating the resulting expression using other integration methods.

In this article, we’ll show you examples of rational functions that will benefit from this technique when we do integrate them. We’ll also take a quick refresher on the general rules when decomposing rational functions to partial functions. You’ll also learn how to apply this technique to integrate different rational functions.

Make sure to have a set of paper or your workbook with you since we’ll be working on different rational functions!

What is a partial fraction integration?

Integration by parts (IBP) is a helpful technique that allows us to integrate functions that can be written as a product of two functions. It helps us integrate complex functions by rearranging the original function so that we’re left with integrals that are easier to work on. Here are some functions that will be much easier to integrate through

\begin{aligned}{\color{Orchid} \int\dfrac{x}{(4 – x)(x + 5)} \phantom{x}dx}\phantom{x}, {\color{DarkGreen} \int\dfrac{1}{x^2 + 4x + 3} \phantom{x}dx} \phantom{x},{\color{DarkOrange} \int\dfrac{x^4}{x^3 – 5x} \phantom{x}dx}\end{aligned}

When we can rewrite complex rational functions as sum or difference of simpler rational functions, it’ll be easier for us to integrate on the decomposed rational functions. These are three examples of rational functions that can be decomposed into partial functions:

Integrating a Rational Function

Decomposing the Rational Function

Integrated Expression

\begin{aligned}\color{Orchid} \int\dfrac{x}{(4 – x)(x + 5)} \phantom{x}dx\end{aligned}

\begin{aligned}\color{Orchid} \int \left(- \dfrac{4}{9(x – 4)}\right)\phantom{x}dx\end{aligned}

\begin{aligned}\color{Orchid} -\dfrac{4}{9}\ln |x – 4| – \dfrac{5}{9}\ln|x + 5|
\end{aligned}

\begin{aligned}\color{DarkGreen} \int\dfrac{1}{x^2 + 4x + 3} \phantom{x}dx\end{aligned}

\begin{aligned}\color{DarkGreen} \int\left[\dfrac{1}{2(x + 1)} -\dfrac{1}{2(x + 3)}\right ] \phantom{x}dx\end{aligned}

\begin{aligned}\color{DarkGreen} \dfrac{1}{2}\ln \left|x+1\right|-\dfrac{1}{2}\ln \left|x+3\right|+C\end{aligned}

\begin{aligned}\color{DarkOrange} \int\dfrac{x^4}{x^3 – 5x} \phantom{x} dx\end{aligned}

\begin{aligned}\color{DarkOrange} \int\left(x +\dfrac{5x}{x^2 – 5}\right )  \phantom{x}dx\end{aligned}

\begin{aligned}\color{DarkOrange} \dfrac{x^2}{2} + \dfrac{5}{2}\ln\left|x^2 – 5\right|+ C\end{aligned}

In the next sections, we’ll show you how to rewrite rational functions by partial fraction decomposition and discuss the process of integrating functions like the three rational functions.

What is the process of partial fraction decomposition?

We’ve learned how to decompose rational functions into partial functions in our Algebra class. When given a rational function, $\dfrac{P(x)}{Q(x)}$ and $Q(x) \neq 0$, we can rewrite this function as $\boldsymbol{\dfrac{P(x)}{Q(x)} = S(x) + \dfrac{R(x)}{Q(x)}}$. This is how we decompose rational functions to simpler rational functions. When decomposing rational functions into partial functions, check for $P(x)$ and $Q(x)$’s degrees:

  • When $P(x)$’s degree is greater than or equal to $Q(x)$’s degree, divide $Q(x)$ by $P(x)$ then take note of the remainder. The divisor becomes $\boldsymbol{S(x)}$ and the remainder becomes $\boldsymbol{R(x)}$.
  • When $P(x)$’s degree is less than $Q(x)$’s degree, we can properly apply the partial composition technique we’ve learned in the past. Focus on the denominator to have an idea of how to rewrite the rational function.

Here’s a table you can use as a guide when decomposing rational functions based on their denominators:

Denominator Form

Partial Fraction Form and

Example

\begin{aligned}ax + b\end{aligned}

\begin{aligned}\dfrac{A }{x + b}\end{aligned}

\begin{aligned}\dfrac{x – 1}{(3x -1)(2x + 1)} = -\dfrac{2}{5\left(3x-1\right)}+  \dfrac{3}{5\left(2x+1\right)}\end{aligned}

\begin{aligned}(ax + b)^n\end{aligned}

\begin{aligned}\dfrac{A}{(x +b)} + \dfrac{B}{(x +b)^2}+ … +\dfrac{Z}{(x +b)^n}   \end{aligned}

\begin{aligned}\dfrac{5x -1}{(x – 1)^4} = \dfrac{5}{\left(x -1\right)^3}+  \dfrac{4}{\left(x -1\right)^4}\end{aligned}

\begin{aligned}ax^2 +bx + c\end{aligned}

\begin{aligned}\dfrac{Ax + B }{ax^2 + bx + c}\end{aligned}

\begin{aligned}\dfrac{2x}{x^2 – 6x + 9} = \dfrac{2}{x-3} + \dfrac{6}{x^2 – 6x +9}\end{aligned}

\begin{aligned}(ax^2 +bx + c)^n \end{aligned}

\begin{aligned}\dfrac{Ax + B }{ax^2 + bx + c} + \dfrac{Cx + D }{(ax^2 + bx + c)^2} + …++ \dfrac{Yx + Z }{(ax^2 + bx + c)^n} \end{aligned}

\begin{aligned}\dfrac{1}{\left(x^2 -4x+ 3\right)^2} =\dfrac{1}{4\left(x-1\right)} +\dfrac{1}{4\left(x-1\right)^2}- \dfrac{1}{4\left(x-3\right)} +\dfrac{1}{4\left(x-3\right)^2}\end{aligned}

Once you have an idea on the partial fractions forms for our rational function, find the coefficients of the partial fractions then rewrite the rational functions into sums or differences of partial fractions. Don’t worry, we’ve prepared enough examples of integrating rational functions for you to practice your knowledge of partial fraction decomposition.

How to solve partial fraction integration?

The challenge in integrating rational functions using partial functions is actually the part where we decompose the expression. The rest of the steps are straightforward, so here are three important pointers to keep in mind when integrating rational functions using this technique:

  • Decompose the rational function into simpler partial fractions.
  • Apply sum or different property for integrals,$\int [f(x) \pm g(x)]\phantom{x}dx = \int f(x)\phantom{x} dx \pm \int g(x)\phantom{x} dx $, to distribute the integral operation.
  • Integrate each partial function using other integration techniques and properties.

The best way to master this technique is through practice. Why don’t we begin by integrating the rational function, $\dfrac{5}{x^2 – 4x + 3}$?

\begin{aligned}\int \dfrac{5}{x^2 – 4x + 3} \phantom{x}dx\end{aligned}

Whenever possible, express the denominator in factored first. We’re looking for partial fractions of the forms, $\dfrac{A}{(x –a)}$ and $\dfrac{B}{(x –b)}$, where $x –a$ and $x –b$ are the factors of the denominator.

\begin{aligned}\dfrac{5}{x^2 – 4x + 3} &= \dfrac{5}{(x -1)(x -3)}\\&= \dfrac{A}{x -1} + \dfrac{B}{x -3}\end{aligned}

Add the two partial fractions in terms of $A$ and $B$, then equate the numerator to $5$.

\begin{aligned}\dfrac{A(x -3) + B(x -1)}{(x- 1))(x -3)}&=\dfrac{5}{(x- 1))(x -3)}\\\\A(x -3) +B(x -1) &= 5\\Ax – 3A + Bx – B &= 5\\ {\color{Orchid}(A +B)}x -{\color{Green}(3A + B)} &= {\color{Orchid}0}x + {\color{Green}5}\end{aligned}

Another way of simplifying this equation is by multiplying both sides of the equation by $x^2 -4x +3$. This means that $\color{Orchid}A + B = 0$ and $\color{Green} -(3A + B) = 5$, so use the two equations to find the values of $A$ and $B$.

\begin{aligned}A + B&= 0 \\A &= -B\\\\(3A + B) &= -5\\-3B + B &= -5\\B &=\dfrac{5}{2}\\A&= -\dfrac{5}{2}\end{aligned}

Rewrite the rational function using these results.

\begin{aligned}\dfrac{5}{x^2 – 4x + 3} &= \dfrac{A}{(x – 1)} + \dfrac{B}{(x – 3)}\\&=\dfrac{-\dfrac{5}{2}}{(x – 1)} + \dfrac{\dfrac{5}{2}}{(x – 3)}\\&= -\dfrac{5}{2(x -1)} + \dfrac{5}{2(x -3)}\\&= -\dfrac{5}{2}\left(\dfrac{1}{x -1} – \dfrac{1}{x -3} \right )\\\\ \int \dfrac{5}{x^2 – 4x + 3}\phantom{x}dx &= \int -\dfrac{5}{2}\left(\dfrac{1}{x -1} – \dfrac{1}{x -3} \right ) \phantom{x}dx\end{aligned}

Now that we have the rational function rewritten as a difference between two partial functions, we can now integrate the function.

  • Factor $-\dfrac{5}{2}$ out of the integral using the constant multiple property of integrals.
  • Apply the difference property of integrals, $\int [f(x) – g(x)]\phantom{x} dx = \int f(x)\phantom{x} dx – \int g(x)\phantom{x} dx$.
  • Use the substitution method to integrate $\dfrac{1}{x -1}$ and $\dfrac{1}{x -3}$.

\begin{aligned}\int \dfrac{5}{x^2 – 4x + 3}\phantom{x}dx &= \int -\dfrac{5}{2}\left(\dfrac{1}{x -1} – \dfrac{1}{x -3} \right ) \phantom{x}dx\\&= -\dfrac{5}{2}\int \left(\dfrac{1}{x -1} – \dfrac{1}{x -3} \right ) \phantom{x}dx\\&= -\dfrac{5}{2} \left(\int\dfrac{1}{x -1}\phantom{x}dx – \int\dfrac{1}{x -3}\phantom{x}dx \right ) \end{aligned}

For the substitution method, assign $u = x -1$ and $v = x -3$. Integrate the resulting expressions using the antiderivative formula, $\int \dfrac{1}{x}\phantom{x}dx = \ln |x| +C$.

\begin{aligned} u &= x- 1\\du &= 1 dx\end{aligned}

\begin{aligned} v &= x- 3\\dv &= 1 dx\end{aligned}

\begin{aligned} \int \dfrac{1}{x -1} \phantom{x}dx &= \int \dfrac{1}{u} \cdot du \phantom{dx}\\&= \ln |u| + C\\&= \ln |x -1| + C\end{aligned}

\begin{aligned} \int \dfrac{1}{x -3} \phantom{x}dx &= \int \dfrac{1}{v} \cdot dv \phantom{dx}\\&= \ln |v| + C\\&= \ln |x -3| + C\end{aligned}

Substitute $\int\dfrac{1}{x -1}\phantom{x}dx$ and $\int\dfrac{1}{x -3}\phantom{x}dx$ with their integrated forms.

\begin{aligned}\int \dfrac{5}{x^2 – 4x + 3}\phantom{x}dx&= -\dfrac{5}{2} \left(\ln|x – 1| – \ln|x -3|\right) + C \end{aligned}

This means that $\int \dfrac{5}{x^2 – 4x + 3} \phantom{x}dx = -\dfrac{5}{2} \left(\ln|x – 1| – \ln|x -3|\right) + C$.

From this example, we can see how it’s now possible for us to integrate complex rational functions by rewriting it to simpler partial functions. We’ve prepared more examples for you to work on. When you feel more confident with this integral technique, head over to our last section to try out more problems!

Example 1

Evaluate the integral, $\int \dfrac{2x^3 – x}{x + 1} \phantom{x}dx$.

Solution

Since the rational function we’re working with has a numerator that is two degrees higher than the denominator, express $\dfrac{2x^3 – x}{x + 1}$ in terms of its quotient and remainder.

 \begin{aligned}\dfrac{2x^3 -x}{x + 1} &= (2x ^2 – 2x + 1) + \dfrac{-1}{x + 1}\\&=  (2x ^2 – 2x + 1) – \dfrac{1}{x + 1}\end{aligned}

With this form, it will now be easier for us to integrate the expression.  Apply the difference property to distribute the integral operation.

\begin{aligned}\int \dfrac{2x^3 – x}{x + 1} \phantom{x}dx &= \int \left[(2x ^2 – 2x + 1) + \dfrac{-1}{x + 1} \right ] \phantom{x}dx\\&=  \int (2x ^2 – 2x + 1) \phantom{x}dx – \int\dfrac{1}{x + 1} \phantom{x}dx\end{aligned}

From a complex expression that we need to integrate, we can now work on simpler functions that we have already learned how to integrate in the past. We’ve broken down the steps for each group to guide you:

 

\begin{aligned}\boldsymbol{\int (2x ^2 – 2x + 1) \phantom{x} dx}\end{aligned}

\begin{aligned}\int (2x ^2 – 2x + 1) \phantom{x} dx &= \int 2x ^2\phantom{x} dx -\int2x\phantom{x} dx + \int1 \phantom{x} dx\\&= 2\int x ^2\phantom{x} dx -2\int x\phantom{x} dx + \int1 \phantom{x} dx \\&= 2 \left(\dfrac{x^{2 + 1}}{2 + 1} \right ) -2 \left(\dfrac{x^{1 + 1}}{1 + 1} \right ) + x + C\\&= \dfrac{2x^3}{3} – x^2 + x +C\end{aligned}

\begin{aligned}\boldsymbol{\int  \dfrac{1}{x -1} \phantom{x} dx}\end{aligned}

\begin{aligned}u &= x -1\\du &=dx\\ \Rightarrow\int \dfrac{1}{x – 1}\phantom{x} dx &=\int \dfrac{1}{u} \phantom{x}du\\&= \ln|u| + C\\&= \ln |x -1| + C\end{aligned}

Let’s go back to our previous expression, $\int (2x ^2 – 2x + 1) \phantom{x}dx – \int\dfrac{1}{x + 1} \phantom{x}dx$, and substitute the expressions we have from the table. Since $C$ represents an arbitrary constant, we can simply use one $C$ for our final expression.

\begin{aligned}\int (2x ^2 – 2x + 1) \phantom{x}dx – \int\dfrac{1}{x + 1} \phantom{x}dx &= \left(\dfrac{2x^3}{3} – x^2 + x\right) + \ln|x -1| + C\\&= \dfrac{2x^3}{3} – x^2 + x + \ln|x -1| + C\end{aligned}

This problem shows how helpful this technique is when integrating complex rational functions.

 

Example 2

Evaluate the integral, $\int \dfrac{x^2 – 4x + 1}{x^3 – 21x + 20} \phantom{x}dx$.

Solution

Since the denominator is one degree higher than the numerator, we can decompose the rational function into simpler partial fractions. When we have a polynomial in the rational function’s denominator, attempt to factor it first before decomposing the partial fraction.

\begin{aligned}x^3 – 21x + 20 &= x^3 – x -20x + 20\\&=x(x^2 – 1) -20(x -1)\\&= x(x -1)(x + 1) – 20(x -1)\\&= (x- 1)(x^2 + x -20)\\&= (x -1)(x -4)(x + 5))\end{aligned}

Since the denominator can be rewritten as three factors of the form, $ax + b$, so we can write $\dfrac{x^2 – 4x + 1}{x^3 – 21x + 20}$ as the sum of three partial fractions. Each partial fraction will have a unique factor assigned from the denominator’s factored form.

\begin{aligned}\dfrac{x^2 -4x + 1}{x^3 – 21x + 20}&= \dfrac{x^2 -4x + 1}{(x- 1)(x -4)(x + 5)}\\ &= \dfrac{A}{x – 1} + \dfrac{B}{x – 4} + \dfrac{C}{x+ 5}\end{aligned}

Multiply both sides of the equation to end up with the equation shown below.

\begin{aligned}x^2 -4x + 1&= \dfrac{x^2 -4x + 1}{(x- 1)(x -4)(x + 5)}\\ &=A(x -4)(x +5) + B(x -1)(x +5) + C(x-1)(x -4)\\&= (A + B+ C)x^2 + (A + 4B – 5C)x+ (-20A – 5B + 4C) \end{aligned}

Equate each corresponding coefficients then solve for $A$, $B$, and $C$.

\begin{aligned}{\color{Orchid}1}x^2 + {\color{Green}-4}x + {\color{DarkOrange}1} &= {\color{Orchid}(A + B+ C)}x^2 + {\color{Green}(A + 4B – 5C)}x+ {\color{DarkOrange}(-20A – 5B + 4C)}\end{aligned}

\begin{aligned}A + B + C &= 1\\A+ 4B – 5C &= -4\\-20A – 5B + 4C &= 1\\\\-18A &= -2\\A&= \dfrac{1}{9}\\B&= \dfrac{1}{27}\\ C&= \dfrac{23}{27} \end{aligned}

Hence, we have $\dfrac{x^2 – 4x + 1}{x^3 – 21x + 20} = \dfrac{1}{9(x – 1)} + \dfrac{1}{27(x – 4)} + \dfrac{23}{27(x+ 5)}$. Rewrite our original integrand using the partial fractions.

\begin{aligned}\int \dfrac{x^2- 4x + 1}{x^3 -21x + 20} \phantom{x}dx &= \int \left[\dfrac{1}{9(x-1)} + \dfrac{1}{27(x-4)} + \dfrac{23}{27(x+ 5)} \right]\phantom{x}dx\\&= \int\dfrac{1}{9(x-1)}\phantom{x}dx +\int\dfrac{1}{27(x-4)}\phantom{x}dx+ \int\dfrac{23}{27(x+ 5)}\phantom{x}dx\end{aligned}

Integrate each group of terms by using the substitution method and the integral property, $\int \dfrac{1}{x}\phantom{x} dx = \ln|x| +C$.

\begin{aligned}\boldsymbol{\int \dfrac{1}{9(x -1)} \phantom{x}dx}\end{aligned}

\begin{aligned}u &= x -1\\du &= dx\\\int \dfrac{1}{9(x- 1)}\phantom{x}dx &= \int \dfrac{1}{9u}\phantom{x}du\\&= \dfrac{1}{9}\int \dfrac{1}{u}\phantom{x}du\\&= \dfrac{1}{9}\ln|u| + C\\&= \dfrac{1}{9}\ln |x -1| + C\end{aligned}

\begin{aligned}\boldsymbol{\int \dfrac{1}{27(x -4)} \phantom{x}dx}\end{aligned}

\begin{aligned}u &= x -4\\du &= dx\\\int \dfrac{1}{27(x- 4)}\phantom{x}dx &= \int \dfrac{1}{27u}\phantom{x}du\\&= \dfrac{1}{27}\int \dfrac{1}{u}\phantom{x}du\\&= \dfrac{1}{27}\ln|u| + C\\&= \dfrac{1}{27}\ln |x -4| + C\end{aligned}

\begin{aligned}\boldsymbol{\int \dfrac{23}{27(x + 5)} \phantom{x}dx}\end{aligned}

\begin{aligned}u &= x +5\\du &= dx\\\int \dfrac{23}{27(x + 5)}\phantom{x}dx &= \int \dfrac{23}{27u}\phantom{x}du\\&= \dfrac{23}{27}\int \dfrac{1}{u}\phantom{x}du\\&= \dfrac{23}{27}\ln|u| + C\\&= \dfrac{23}{27}\ln |x + 5| + C\end{aligned}

Let’s use these expressions to integrate our original expression.

\begin{aligned}\int \dfrac{x^2- 4x + 1}{x^3 -21x + 20} \phantom{x}dx &= \int\dfrac{1}{9(x-1)}\phantom{x}dx +\int\dfrac{1}{27(x-4)}\phantom{x}dx+ \int\dfrac{23}{27(x+ 5)}\phantom{x}dx\\&= \dfrac{1}{9}\ln |x -1| + \dfrac{1}{27}\ln |x -4|+ \dfrac{23}{27}\ln |x + 5| + C\end{aligned}

Hence, through partial fraction decomposition, we have $\int \dfrac{x^2 – 4x + 1}{x^3 – 21x + 20} \phantom{x}dx = \dfrac{1}{9}\ln |x -1| + \dfrac{1}{27}\ln |x -4|+ \dfrac{23}{27}\ln |x + 5| + C$.

 Practice Questions

1. Evaluate the following integrals:
a. $\int \dfrac{4x^3 – x^2 + 1}{x^2 – 2}\phantom{x}dx$
b. $\int \dfrac{x^2 + 1}{x + 4}\phantom{x}dx$
c. $\int \dfrac{x^3 – 4x}{x^2 + 3}\phantom{x}dx$

2. Evaluate the following integrals:
a. $\int \dfrac{x -4}{x^2 – 2x + 1}\phantom{x}dx$
b. $\int \dfrac{x^2}{x^3 – 1}\phantom{x}dx$
c. $\int \dfrac{x^2 + 4x + 4}{x^3 -10x^2 + 19x+ 30}\phantom{x}dx$

3. Evaluate the following integrals:
a. $\int \dfrac{1}{(x -1)^2(x + 2)} \phantom{x}dx$
b. $\int \dfrac{1}{(x+ a)(x +b)} \phantom{x}dx$
c. $\int \dfrac{ax}{x^2 – bx}\phantom{x}dx$

Answer Key

1.
a. $\dfrac{3x^2}{2}+ 6x – 18 + 13 \ln|x -2| +C$
b. $\dfrac{x^2}{2} -4x – 24+ 17\ln \left|x+4\right|+C$

c. $\dfrac{1}{2}x^2+ \dfrac{3}{2} -\dfrac{7}{2} \ln \left|x^2+3\right| +C$
2.
a. $\dfrac{3}{x-1} + \ln \left|x-1\right| + C$

b. $\dfrac{1}{3}\ln \left|x^3 – 1\right| +C$
c. $\dfrac{1}{42}\ln \left|x+1\right| – \dfrac{49}{6}\ln \left|x – 5\right| +\dfrac{64}{7}\ln \left|x-6\right| +C$

3.
a. $\dfrac{1}{9}\ln \left|x + 2\right| -\dfrac{1}{9}\ln \left|x- 1\right| – \dfrac{1}{3\left(x -1\right)} +C$
b. $\dfrac{1}{a –b}(\ln |b + x| – \ln |a + x|) +C$
c. $a \ln|b – a| + C$

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