### Video Transcript

In this video, we’ll learn how to use partial fractions to evaluate integrals of rational functions with linear factors. It’s likely that you will have worked extensively with algebraic fractions, multiplying, dividing, adding, subtracting, and simplifying them. Hopefully, by this stage, you’re also confident with the process that’s required for integrating simple reciprocal functions, integration by substitution and by parts.

We’ll now look to extend these ideas and learn how splitting a fraction into fractions with linear denominators can make the process of integrating a ratio of polynomials much simpler. A single fraction with two distinct linear factors in the denominator can be split into two separate fractions with linear denominators. This is known as splitting it or decomposing it into partial fractions. And at this level, partial fractions are mainly used for binomial expansions and integration.

To illustrate the method, let’s remind ourselves how we find the sum of two fractions with linear denominators. We multiply the numerator and the denominator of each fraction by the denominator of the other fraction, thereby creating a common denominator and equivalent fractions. Here that’s three times 𝑥 plus five plus two times 𝑥 plus one over 𝑥 plus one times 𝑥 plus five. And distributing our parentheses, and we see that we’re left with five 𝑥 plus 17 over 𝑥 squared plus six 𝑥 plus five.

We can now see that if we were asked to integrate five 𝑥 plus 17 over 𝑥 squared plus six 𝑥 plus five with respect to 𝑥, we might struggle. But we now know that we can write this as three over 𝑥 plus one plus two over 𝑥 plus five. And then we recall the processes for integrating simple reciprocal functions. The integral of one over 𝑥 plus 𝑎 for real constant 𝑎 is equal to the natural log of the absolute value of 𝑥 plus 𝑎 plus some constant of integration 𝑐.

And we can now see that the integral of five 𝑥 plus 17 over 𝑥 squared plus six 𝑥 plus five is three times the natural log of the absolute value of 𝑥 plus one plus two times the natural log of the absolute value of 𝑥 plus five plus 𝑐. Well, that was all fine and well. But how do we figure out what our partial fractions are?

Imagine we wanted to evaluate this indefinite integral. By simply working backwards from the example we just did, we see that we can split this into partial fractions, whose respective denominators are 𝑥 plus six and 𝑥 minus one. We’ll call their numerators 𝐴 and 𝐵. There are two ways to find the constants 𝐴 and 𝐵. Those are substitution and equating coefficients. We’ll also need to know what to do in the situation where we have an improper or a top heavy fraction. But let’s just begin with a simple example.

Use partial fractions to evaluate the indefinite integral of 𝑥 plus four over 𝑥 plus six times 𝑥 minus one with respect to 𝑥.

Remember, we need to rewrite this integrand using partial fractions. So we begin by reversing the process we would take when adding algebraic fractions. We write it as 𝐴 over 𝑥 plus six plus 𝐵 over 𝑥 minus one. And there are two ways for us to work out the constants 𝐴 and 𝐵. They’re substitution and equating coefficients.

Let’s begin by looking at the substitution method. Let’s imagine we’re adding these fractions. We multiply the numerator and denominator of the first fraction by 𝑥 minus one and the numerator and denominator of the second fraction by 𝑥 plus six. So we see that 𝑥 plus four over 𝑥 plus six times 𝑥 minus one is equal to 𝐴 times 𝑥 minus one plus 𝐵 times 𝑥 plus six all over 𝑥 plus six times 𝑥 minus one.

Notice that the denominator of the fractions on both sides of our equation are equal. This means that, for the fractions themselves to be equal, their numerators must be. And we can say that 𝑥 plus four is equal to 𝐴 times 𝑥 minus one plus 𝐵 times 𝑥 plus six. Okay, so far, so good.

Now we want to find a way to eliminate one of the constants from this equation. Well, we see that if we set 𝑥 to be equal to one, this bit here becomes 𝐴 times one minus one, which is 𝐴 times zero, which is zero. So let’s set 𝑥 be equal to one. When we do, we see that one plus four is equal to 𝐴 times zero plus 𝐵 times one plus six, which simplifying gives us five equals seven 𝐵. And now we have an equation in 𝐵.

So we can solve this by dividing both sides by seven. And we see that 𝐵 is equal to five-sevenths. Brilliant, so let’s repeat this process to help us establish the value of 𝐴. And you might wish to pause the video for a moment and think about what substitution would eliminate 𝐵 from this equation.

If we let 𝑥 be equal to negative six, then the second term over here becomes 𝐵 times zero. So the 𝐵’s going to be eliminated. And if we let 𝑥 be equal to negative six, our entire equation becomes negative six plus four equals 𝐴 times negative six minus one plus 𝐵 times zero, which simplifies to negative two equals negative seven 𝐴. Now we have an equation in 𝐴. And dividing both sides by negative seven, we obtain 𝐴 to be equal to two-sevenths. And we’ve successfully decomposed into partial fractions.

We can say that 𝑥 plus four over 𝑥 plus six times 𝑥 minus one is equal to two over seven times 𝑥 plus six plus five over seven times 𝑥 minus one. And we’re now of course able to integrate our expression with respect to 𝑥. Remembering that the integral of the sum of functions is the same as the sum of the integrals of those respective functions. And of course, we can take out any constant factors. And we see that our integral is equal to two-sevenths of the integral of one over 𝑥 plus six d𝑥 plus five-sevenths of the integral of one over 𝑥 minus one d𝑥.

Well, the integral of one over 𝑥 plus six is the natural log of the absolute value of 𝑥 plus six. And the integral of one over 𝑥 minus one is the natural log of the absolute value of 𝑥 minus one. And of course, since this is an indefinite integral, we must add that constant 𝑐. And we’re done. The indefinite integral of 𝑥 plus four over 𝑥 plus six times 𝑥 minus one with respect to 𝑥 is two-sevenths times the natural log of the absolute value of 𝑥 plus six plus five-sevenths of the natural log of the absolute value of 𝑥 minus one plus the constant 𝑐.

We are now going to consider how we could’ve got here using the method of equating coefficients. The starting process is the same. We need to get to this stage here. We want to distribute the parentheses on the right-hand side. And when we do, we see that 𝑥 plus four is equal to 𝐴𝑥 minus 𝐴 plus 𝐵𝑥 plus six 𝐵.

Now this next step isn’t entirely necessary. But it can help us figure out what to do next. We collect together like terms. And we see that 𝑥 plus four is equal to 𝐴 plus 𝐵 times 𝑥 plus negative 𝐴 plus six 𝐵 or six 𝐵 minus 𝐴. And now we have two families of terms, if you will. We have 𝑥 to the power of ones, and then we have these constants. And we can say that those are 𝑥 to the power of zeros.

We want to equate coefficients for these terms. Let’s begin by equating the coefficients of 𝑥 to the power of zero, or just the constants. On the left-hand side, we have four. And on the right-hand side, we have six 𝐵 minus 𝐴. Next, we’ll equate the coefficients of 𝑥 to the power of one. The coefficient of 𝑥 on the left-hand side is one. And on the right-hand side, that’s 𝐴 plus 𝐵. So we now have a pair of simultaneous equations which we can begin to solve by first adding.

Negative 𝐴 plus 𝐴 is zero. So we see that when we add our pair of simultaneous equations, we end up with five equals seven 𝐵. And solving this equation for 𝐵, we find that 𝐵 is equal to five-sevenths. We’re gonna then substitute this value of 𝐵 into either of our original equations. I’m going to choose this one here. So one is equal to 𝐴 plus five-sevenths. Then subtracting five-sevenths from both sides, we obtain 𝐴 to be equal to two-sevenths. And the rest of the process is exactly the same.

We have our partial fractions, and we can integrate each of them. And we obtain the indefinite integral to be equal to two-sevenths of the natural log of 𝑥 plus six plus five-sevenths of the natural log of 𝑥 minus one plus 𝑐. Both of these methods are equally as valid as one another. And there may even be times where you’ll need to combine the methods, which is also fine. We’ll now have a look at an example which involves three linear terms.

Use partial fractions to evaluate the indefinite integral of one over 𝑡 cubed plus 𝑡 squared minus two 𝑡 with respect to 𝑡.

When rewriting the integrand using partial fractions, we’ll be looking to reverse the process we would take when adding algebraic fractions. Before we can do this though, we need to decide what the denominator of these fractions might be. So let’s see if we can factor the denominator.

It’s clearly a cubic, but there’s a common factor of 𝑡. So we’ll factor the denominator by 𝑡 to get 𝑡 times 𝑡 squared plus 𝑡 minus two. We can then factor this quadratic expression the usual way. 𝑡 squared plus 𝑡 minus two is equal to 𝑡 minus one times 𝑡 plus two. We can now split this into partial fractions. These are 𝐴 over 𝑡 plus some other constant 𝐵 over 𝑡 minus one plus 𝐶 over 𝑡 plus two.

Next, we’re going to add these three fractions. We’ll need to multiply the numerator and denominator of each of them by the denominator of the other two. So the numerator becomes 𝐴 times 𝑡 minus one times 𝑡 plus two plus 𝐵 times 𝑡 times 𝑡 plus two plus 𝐶 times 𝑡 times 𝑡 minus one. And now we notice that the denominator of the fractions on both sides of this equation are equal. And for the fractions themselves to be equal, this means in turn that their numerators must be equal. So we can say that one is equal to 𝐴 times 𝑡 minus one times 𝑡 plus two plus 𝐵𝑡 times 𝑡 plus two plus 𝐶𝑡 times 𝑡 minus one. We’re going to use the method of substitution to help us find the values for 𝐴, 𝐵, and 𝐶.

Notice that if we let 𝑡 be equal to zero, we can instantly eliminate 𝐵 and 𝐶. So we substitute 𝑡 equals zero into our equation. And we find that one is equal to 𝐴 times zero minus one times zero plus two, which simplifies to one equals negative two 𝐴. Dividing both sides of this equation by negative two, and we obtain 𝐴 to be equal to negative one-half.

Next, we spot that if we let 𝑡 be equal to one, we’re going to eliminate 𝐴 and 𝐶. Substituting 𝑡 equals one gives us one is equal to three 𝐵. And when we divide by three, we see that 𝐵 is equal to one-third. Finally, we’ll let 𝑡 be equal to negative two. This will eliminate 𝐴 and 𝐵 this time. This gives us one is equal to six 𝐶. And dividing through by six, we find 𝐶 to be equal to one-sixth.

Let’s clear some space for the next step. We can say that one over 𝑡 times 𝑡 minus one times 𝑡 plus two is equal to negative one over two 𝑡 plus one over three times 𝑡 minus one plus one over six times 𝑡 plus two. We’re now ready to integrate with respect to 𝑡. The integral of one over 𝑡 is the natural log of the absolute value of 𝑡. So our first term integrates to negative a half times the natural log of 𝑡. The interval of one over three times 𝑡 minus one is a third times the natural log of the absolute value of 𝑡 minus one. And the integral of one over six times 𝑡 plus two is equal to a sixth times the natural log of the absolute value of 𝑡 plus two. We mustn’t forget since we’re dealing with an indefinite integral to add that constant of integration 𝐶.

Now this method is fab. But it’s usually important to know that it can’t be used when there’s a repeated factor in the denominator. Let’s see what we’ll do in this case.

Use partial fractions to find an analytic expression for the integral of three 𝑡 squared minus nine 𝑡 plus eight over 𝑡 times 𝑡 minus two squared, evaluated with respect to 𝑡 and between the limits of one and 𝑥.

Notice here that we have a repeated factor in the denominator of our fraction. A single fraction which has a repeated linear factor can be split into two or more separate fractions. But there is a special method for dealing with this linear factor. We list the repeated factor using increasing powers. We’re going to rewrite our integrand as 𝐴 over 𝑡 plus 𝐵 over 𝑡 minus two plus 𝐶 over 𝑡 minus two all squared.

Next, we’re going to add these fractions, remembering that the denominator is going to be 𝑡 times 𝑡 minus two all squared. So we multiply the numerator and denominator of our first fraction by 𝑡 minus two squared. For our second fraction, we multiply it by 𝑡 times 𝑡 minus two. And for our third fraction, we multiply that by simply 𝑡. And we see that the numerator is now 𝐴 times 𝑡 minus two all squared plus 𝐵𝑡 times 𝑡 minus two plus 𝐶𝑡.

We can see that since these expressions are equivalent, three 𝑡 squared minus nine 𝑡 plus eight must be equal to 𝐴 times 𝑡 minus two squared plus 𝐵𝑡 times 𝑡 minus two plus 𝐶𝑡. And we’ll use the method of substitution to find the values for 𝐴, 𝐵, and 𝐶. We’ll begin by letting 𝑡 be equal to two, with the aim of eliminating 𝐴 and 𝐵. On the left-hand side, we have three times two squared minus nine times two plus eight, which is two. And on the right-hand side, we have two 𝐶. Dividing through by two, and we find that 𝐶 is equal to one.

Next, we’ll let 𝑡 be equal to zero. This time, that will eliminate 𝐵 and 𝐶. Substituting 𝑡 equals zero, we get eight equals four 𝐴. And then when we divide through by four, we find that 𝐴 is equal to two. The problem is, we’ve now run out of substitutions to make. So we’re going to use equating coefficients to find the value of 𝐵. So we distribute our parentheses. We’re going to equate the coefficients of 𝑡 squared.

We have 𝐴 and 𝐵 on the right-hand side and three on the left. So we can say that three must be equal to 𝐴 plus 𝐵. We already established though that 𝐴 was equal to two. So we can substitute 𝐴 is equal to two and say that three is equal to two plus 𝐵. And subtracting two from both sides, we find that 𝐵 is equal to one.

Let’s clear some space for the next step. We now see that our integrand is equal to two over 𝑡 plus one over 𝑡 minus two plus one over 𝑡 minus two squared. We’re going to integrate this between the limits of one and 𝑥. The integral of two over 𝑡 is two times the natural log of 𝑡. The integral of one over 𝑡 minus two is the natural log of 𝑡 minus two. And then we can use the reverse chain rule on 𝑡 minus two to the power of negative two. And we find that the integral of that is negative one over 𝑡 minus two.

We’re now going to substitute these limits in. When we do, we find this is equal to two times the natural log of the absolute value of 𝑥 plus the natural log of the absolute value of 𝑥 minus two minus one over 𝑥 minus two minus one. This example is a nice example of how a mixture of substitution and equating coefficients can help us to achieve the solution. In our final example, we’ll consider what happens when we’re working with an improper fraction.

Express 𝑥 cubed over 𝑥 squared plus two 𝑥 plus one in partial fraction form.

Notice that the degree of our numerator is higher than the degree of the denominator. On the numerator, we have 𝑥 cubed. And on the denominator, we only go as far as 𝑥 squared. This means that we know that we have an improper fraction. And we’re going to need to perform polynomial long division. You might wish to pause the video and divide 𝑥 cubed by 𝑥 squared plus two 𝑥 plus one yourself.

When we do, we find that 𝑥 cubed divided by 𝑥 squared plus two 𝑥 plus one is 𝑥 minus two with a remainder of three 𝑥 plus two, which means we can rewrite it as shown. We now have a mixed fraction. Factoring the denominator of the remainder part, and we notice we then have a repeated root. So when writing in impartial fraction form, we write it as 𝐴 over 𝑥 plus one plus 𝐵 over 𝑥 plus one squared.

When we add the fractions, we multiply the first one by 𝑥 plus one. So we have 𝐴 times 𝑥 plus one. And we don’t need to do anything with the second. So we find that, in the numerators, we have three 𝑥 plus two equals 𝐴 times 𝑥 plus one plus 𝐵. We’ll begin by equating coefficients of 𝑥 to the power of one. On the left-hand side, that’s three. And on the right-hand side, that’s 𝐴. So we see that 𝐴 is equal to three.

We then equate coefficients of 𝑥 to the power of zero or constants. And we find that two is equal to 𝐴 plus 𝐵. But we know that 𝐴 is equal to three. So two is equal to three plus 𝐵. We then subtract three from both sides, and we find 𝐵 to be equal to negative one. And we have expressed our fraction in partial fraction form. It’s 𝑥 minus two plus three over 𝑥 plus one minus one over 𝑥 plus one squared. And if we’re so required, we could now integrate it.

In this video, we’ve seen that a single fraction with two distinct linear factors in its denominator can be split into two separate fractions with linear denominators. This is called splitting into partial fractions. And we saw that this can make the integration process considerably easier. We also saw that this method can be applied when there are more than two distinct linear factors. But that a single fraction with a repeated linear factor needs a special rule whereby we list the repeated power in ascending powers.

Finally, we learned that if the fraction is improper, its numerator has a degree equal to or larger than the denominator. And we need to convert it to a mixed fraction by using polynomial long division before it can be expressed as partial fractions.