Partial Fractions

Since the degree of the denominator is higher than the degree of the numerator, we don’t have to perform long division before we start. Instead, we can move straight to factoring the denominator, as follows.

We can see that our denominator is a product of distinct linear functions because our three factors, (x-1), (x+1), and (x-2), are all different.

Once we have it factored, we set our fraction equal to the sum of its component parts, assigning new variables to the numerator of each of our fractions. Since our denominator can be broken down into three different factors, we need three variables, A, B and C to go on top of each one of our new fractions, like so:

Now that we’ve separated our original function into its partial fractions, we multiply both sides by the denominator of the left-hand side. The denominator will cancel on the left-hand side, and on the right, each of the three partial fractions will end up multiplied by all the factors other than the one that was previously included in its denominator.

Our next step is to multiply out all of these terms.

Now we collect like terms together, meaning that we re-order them, putting all the x^2 terms next to each other, all the x terms next to each other, and then all the constants next to each other.

Finally, we factor out the x^2 and the x.

Doing this allows us to equate coefficients on the left and right sides. Do you see how the coefficient on the x^2 term on the left-hand side of the equation is 1? Well, the coefficient on the x^2 term on the right-hand side is (A+B+C), which means those two must be equal. We can do the same for the x term, as well as the constants. We get the following three equations:

Now that we have these equations, our goal will be to solve for each of the three variables. This can easily get confusing, but with practice, you should get the hang of it. If you have one equation with only two variables instead of all three, like our second equation, that’s a good place to start. Adding A to both sides of our second equation gives us

Now we’ll plug in 2A everywhere we find B in our first and third equations and simplify.

Finally, if we add these two equations together, our Cs will cancel and we can solve for A. Depending on your equations, you may have to multiply both sides of one equation by some constant before adding or subtracting will cancel a variable for you.

Once we find the value of one variable, luckily it becomes much easier to determine the values of the others. For example, remember that B=2A. Plugging in our known value for A lets us quickly solve for the value of B.

Of course, knowing A and B, we now plug them into our first equation, A+B+C=1, and determine that the value of C is

Having solved for the values of our three variables, we’re finally ready to plug them back into our partial fractions decomposition. Doing so should produce something that’s easier for us to integrate than our function was originally.

Remembering that the integral of 1/x is ln|x|+C, we’ll get

Let’s move now to the second of our four case types above, in which the denominator will be a product of linear functions, some of which are repeated.

You’ll see that we need to carry out long division before we start factoring, since the degree of the numerator is five, which is greater than the degree of the denominator, which is 4.

Now that the degree of the remainder is less than the degree of the original denominator, we can write our function as

Integrating the 2x-3 will be simple, so for now, let’s focus on the fraction. We’ll factor the denominator.

Given the factors involved in our denominator, you might think that the partial fractions decomposition would look like this

However, the fact that we’re dealing with repeated factors, (x-1 is a factor twice and x+1 is a factor twice), our partial fractions decomposition is actually the following:

When you have repeated factors, you have to include each degree of the factor that is of a degree lesser than or equal to the factor in your partial fractions decomposition. To use another, simpler example, if you were trying to perform a partial fractions decomposition on 1/x^3, your decomposition would be

Let’s continue with our example.

We’ll multiply both sides of our equation by the denominator on the left side, (x-1)^2(x+1)^2, which will cancel the denominator on the left and one of each of the four denominator terms on the right side.

To simplify, we’ll start multiplying all terms on the right side together.

Now we’ll group like terms together.

Equating coefficients on both sides of the equation gives us the following equations.

Now we’ll start solving for variables. If we subtract A from both sides of the first equation, A+C=9, we’ll get C=9-A. If we now plug in 9-A for C into our other three equations, we have:

And simplifying, we get the following:

Let’s now solve our second equation above for B.

Now we’ll plug D-1 into our first (2A+B+D=6) and third (-2A+B+D=7) equations above for B.

Simplifying both equations gives:

Now let’s take the second equation, -2A+2D=8, and solve it for D.

Finally, let’s take 4+A and plug it into our first equation, 2A+2D=7, for D. Then we can simplify and solve for A.

At last! We’ve solved for one variable. Now it’s pretty quick to find the other three. Since we already know the value of A, we can use the equation D=4+A to solve for D.

Moving back one more step, we can use our value for D and the equation from earlier, B=D-1, to solve for B.

And last but not least, going all the way back to our original set of equations, we find the equation C=9-A which, along with our value for A, will allow us to easily find C.

We’ll now take our values for A, B, C and D, and bring back the 2x+3 that we put aside following the long division earlier in this example, and write out our partial fractions decomposition.

Now we can integrate. Using the rule from algebra that 1/x^n=x^-n, we’ll flip the second and fourth terms of the partial fractions decomposition so that they are easier to integrate.

Now that we’ve simplified, we’ll integrate.

At this point, you can try simplifying further, or just leave your answer as is.

Now let’s take a look at an example in which the denominator is a product of distinct quadratic factors. When this is the case, you can represent your function as the product of factors of the form

In order to integrate these terms, you’ll also need the following formula:

So here’s our example for this section:

As always, the first thing to notice is that the degree of the denominator is larger than the degree of the numerator, which means that we don’t have to perform long division before we can start factoring the denominator. So let’s get right to it.

Since we cannot factor the denominator any further, using the first formula above we can write the partial fraction decomposition as

Multiplying both sides of this equation by the denominator on the left gives

Collecting like terms together gives

Equating coefficients on both sides gives us the following equations:

Solving the first equation for A looks like this:

Plugging 1-B into the third equation for A leaves us with the following two equations.

Simplifying these leaves us with

Now if we solve the first of these equations for C, we’ll get

Plugging this into our second equation gives

Now that we’ve solved for the value of B, we can use the equation from just above, C=B-2, to solve for C.

To find A, we’ll again use our value for B, as well as an equation from earlier, A=1-B.

Now that we have all three, we can take these values and plug them back into our partial fractions decomposition, and then integrate.

To integrate the second term, we’ll need to use u-substitution.

Now that we’ve performed u-substitution on our second integral, integrating both the first and second integrals will be easy. We’ll need to use the formula from above to integrate the third term.

In our case, x=x and a=3. M=1 and N=3.

Plugging back in for u and simplifying this whole thing gives us our final answer.

You can either simplify further, or leave this as is.

Last but not least, let’s take a look at an example in which the denominator is a product of quadratic functions, at least some of which are repeated.

We’ll be using the same formulas we did for the last example. I’ll put them here again so we don’t have to go back.

The problem we’ll use for this section is

Remember, when we’re dealing with repeated factors, we have to include every lesser degree of that factor in our partial fractions decomposition, which will be

Multiplying both sides by the denominator of the left-hand side gives us

Multiplying all terms together will leave us with

Grouping like terms together gives us

Now we equate coefficients and write down the equations we’ll use to solve for each of our variables.

Since the second and fifth equations give us the values of C and A, we’ll plug those values into our other three equations first.

Simplifying these should allow us to solve for more variables.

At this point, the only variable we haven’t solved for is D, but if we plug into the second equation the value that we just got for B, we can see that

If we plug all five values back into our partial fractions decomposition, we’ll get

We can now separate our terms into separate integrals, like so:

Let’s use u-substitution on the second integral above.

Our problem therefore becomes

Cancelling the x in the second integral gives

Let’s also use u-substitution for the fourth integral above, but let’s use the letter k instead of u, so that we don’t get confused.

Plugging back in and simplifying will give us a set of integrals we can finally integrate.