# Integration by Partial Fractions

## Introduction to Integration by Partial Fractions

Integration by Partial Fraction Decomposition is a procedure to “decompose” a proper Rational Function (one with a variable in the denominator) into simpler rational functions that are more easily integrated. Typically, we are breaking up one “complicated” fraction into several different “less complicated” fractions. You may have learned how to use this technique in your Algebra class, and it’s quite useful in Calculus! Note that Integration by Partial Fractions is used when U-Substitution Integration (also known as the reverse chain rule or change of variables) doesn’t work easily; a rule of thumb is to try U-sub first!

(Note that Partial Fraction Decomposition may really only be used with rational functions with denominators that factor well. In more advanced cases, you may need to use Completing the Square with Quadratics or another method to get the rational in the right form).

The basic idea is to factor the denominator (if it isn’t already factored) of the complicated factor, and then break it up into different fractions with denominators of those factors. We’ll see that when we do this, we’ll usually end up with Logarithmic Integration or Inverse Trig Function Integration.

If an integral is improper (the degree of the numerator is greater than or equal to the degree of the denominator), use Polynomial Long Division to get a term that’s not a rational function, and then decompose the remaining terms; we’ll show an example below.

## Basic Partial Fraction Decomposition Rules

Here are the basic Partial Fraction Rules when for decomposing fractions, without solving yet for the variables. Notice how combinations of rules can be used, depending on the types of denominators. Remember that these rules apply to rationals that aren’t improper; they apply only to fractions whose denominator has a degree larger than that of the numerator.

This looks really complicated, but it’s not that bad; let’s do some examples.

## Integration by Partial Fractions Examples

Note that you can check these using Wolfram Alpha: just type in something like “Integrate (2x+10)/(12x^2+x-1)”.

Integrate $\displaystyle \int{{\frac{{2x+1}}{{12{{x}^{2}}+x-1}}dx}}$  (first example above):

Since it looks pretty impossible to integrate with U-sub, try to divide it into two separate fractions by factoring the denominator, $\displaystyle \int{{\frac{{2x+1}}{{12{{x}^{2}}+x-1}}dx}}=\int{{\frac{{2x+1}}{{\left( {3x+1} \right)\left( {4x-1} \right)}}dx}}$. (When doing these types of problems in school, the problem will probably be easy to factor; in the “real world,” this would probably not be the case.)

Try to some values A and B where $\displaystyle \frac{{2x+1}}{{\left( {3x+1} \right)\left( {4x-1} \right)}}=\frac{A}{{\left( {3x+1} \right)}}+\frac{B}{{\left( {4x-1} \right)}}$, since then we could integrate log functions. To find what $A$ and $B$ are, multiply each term by a common denominator (you can also think of it as multiplying on top by “what’s missing” on bottom) so we can get rid of fractions:

There are two different ways to solve for $A$ and $B$:

1. Multiply out the last equation and solve for $A$ and $B$ by equating coefficients and setting up a System of Equations to solve:

1. (Usually, the preferred method for linear factors in denominator; it tends to be easier, but it doesn’t always find all the coefficients!) In this method, we don’t even have to distribute the $A$ and $B$. Since setting $\displaystyle x=\frac{1}{4}$ and $\displaystyle x=-\frac{1}{3}$ turns coefficients for $A$ and $B$, respectively, into 0, set $x$ to these values and solve:

Note: this seems a little counter-intuitive, since it appears that we are setting denominators to 0, which is a no-no. But in actuality, we are equating the numerators and ignoring the denominators when using this method, so it works!

Both methods: substitute the $A$ and $B$ back in the equation and integrate:

Let’s try one that’s a little bit more complicated.

Integrate $\displaystyle \int{{\frac{{{{x}^{2}}+2x+2}}{{{{x}^{3}}-3{{x}^{2}}+2x}}dx}}$:

Start by factoring the denominator and getting rid of it by multiplying by a common denominator (you can also think of multiplying the numerator by “what’s missing” in the denominator):

Now set $x$ to values that make the terms above 0:

Substitute the $A$ and $B$ back in the equation and integrate:

## Integration by Partial Fractions with Improper Fractions

Let’s try another one, but one where we have to perform some long division first:

Integrate $\displaystyle \int{{\frac{{{{x}^{3}}+2{{x}^{2}}+3x+7}}{{{{x}^{2}}-x-6}}dx}}$:

Since the degree on the top (3) is greater than the degree of the bottom (2), perform Polynomial Long Division to obtain polynomial and rational functions:

$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,+\,\,3\,\,+\,\,\frac{{12x+25}}{{{{x}^{2}}-x-6}}\\{{x}^{2}}-x-6\,\overline{\left){{{{x}^{3}}+2{{x}^{2}}+3x+7}}\right.}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{{{x}^{3}}-\,\,{{x}^{2}}\,-6x}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3{{x}^{2}}+9x\,+\,7\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,3{{x}^{2}}-\,3x-18}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12x+25\end{array}$

We end up with

Use Partial Fraction Decomposition for the last term:

Now set $x$ to values that make the terms above 0:

Substitute the $A$ and $B$ back in the equation and integrate:

## Example of Rational Function where Partial Fractions are not Needed

Note that not all rational functions need to be decomposed by Partial Fractions. Here is an example:

Integrate $\displaystyle \int{{\frac{{{{x}^{2}}-1}}{{{{x}^{3}}-3x+9}}}}\,dx$:

See how we can use U-Substitution Integration?

## Integration by Partial Fractions with Higher Degrees

This isn’t usually taught in an AB Calculus class, but it gets a little more complicated when we are decomposing rational functions with repeatable factors in the denominator, and factors with higher degrees, which we’ll see below.

As in the chart above, with repeatable factors in the denominator, the rational may be decomposed in a number of ways, with various factors of the numerator. We must account for all these ways (even if some numerators are 0) by using denominators with different powers of this factor.

For example, decompose $\displaystyle \frac{{{{x}^{3}}-10{{x}^{2}}+43x-54}}{{x{{{\left( {x-3} \right)}}^{3}}}}$.

We could set up the partial fraction decomposition this way:

When solving for $A$, $B$, $C$, and $D$, using $x=0$ and $x=3$, you may only get some of these values; you could then plug in random $x$=values to create a system of equations to get the other values. But we’ll just multiply by common denominator, multiply out to equate coefficients, and then use systems to get the coefficients:

The other type of rational function that is more advanced is when we have a higher degree (2 and up) in a prime (unfactorable) factor in the denominator. For example, for a quadratic factor in the denominator that can’t be factored, we’ll have to put a linear factor in the numerator (like $Ax+B$). Here’s one to try:

Decompose (don’t integrate): $\displaystyle \frac{{7{{x}^{2}}-4x+30}}{{\left( {x-3} \right)\left( {{{x}^{2}}+3x+9} \right)}}$

Again, to get the system of equations, multiply everything out and set coefficients together:

Let’s do this one two ways to show we’ll get the same answer:

Decompose (don’t integrate): $\displaystyle \frac{{5x-3}}{{\left( {x-1} \right)\left( {x+1} \right)\left( {x+3} \right)}}$

Decompose (don’t integrate): $\displaystyle \frac{{5x-3}}{{\left( {{{x}^{2}}-1} \right)\left( {x+3} \right)}}$ (same as above)

This is equivalent to the decomposition above!

Here’s one more to set up: decompose $\displaystyle \frac{{5{{x}^{3}}+3{{x}^{2}}+x+5}}{{x{{{\left( {{{x}^{2}}-3} \right)}}^{3}}}}$.

We could set up the partial fraction decomposition this way:

I’m not going to try to solve this one, but we multiply everything out and set coefficients together, and solve a system of equations. We could also or first start with $x=3$ and $x=0$, simplify, use random $x$-values that are the same across the board, and solve the system.

And it turns out that these types of partial fraction integration problems usually result in the Integration of Inverse Trigonometric Functions.

For Practice: Use the Mathway widget below to try an Integration by Partial Fractions problem. Click on Submit (the blue arrow to the right of the problem) and click on Describe the Transformation to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

Learn these rules, and practice, practice, practice!

On to Trigonometric Substitution for Evaluating Integrals  – you are ready!

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