Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ π‘₯⁴/(π‘₯Β² βˆ’ 1) dπ‘₯.

04:59

Video Transcript

Use partial fractions to evaluate the indefinite integral of π‘₯ to the fourth power of π‘₯ squared minus one with respect to π‘₯.

Now, we are actually told how we’re going to evaluate this integral. We need to rewrite our integrand β€” that’s π‘₯ to the fourth power over π‘₯ squared minus one β€” in partial fraction form. Before we do though, we need to notice that the highest power of π‘₯ in the numerator is larger than the highest power of π‘₯ in the denominator. And that means π‘₯ to the fourth power over π‘₯ squared minus one is a top-heavy or an improper fraction. And so, before we can write in partial fraction form, we’re going to use polynomial long division to divide π‘₯ to the fourth power by π‘₯ squared minus one.

Now, it can be really useful to write π‘₯ to the fourth power as π‘₯ to the fourth power plus zero π‘₯ cubed plus zero π‘₯ squared and so on. It just makes the whole process of polynomial long division a little bit easier. So the first thing we do is divide π‘₯ to the fourth power by π‘₯ squared. That’s π‘₯ squared. We then multiply π‘₯ squared by π‘₯ squared and π‘₯ squared by negative one. That gives us π‘₯ to the fourth power minus π‘₯ squared. And of course, in this term, we have zero π‘₯ cubed.

We now subtract these three terms. π‘₯ to the fourth power minus π‘₯ to the fourth power is zero. Zero π‘₯ cubed minus zero π‘₯ cubed is zero. And zero π‘₯ squared minus negative π‘₯ squared is π‘₯ squared. We bring down the remaining terms. And now, we divide π‘₯ squared by π‘₯ squared, which is simply one. This time, we multiply one by each term in our divisor. That’s π‘₯ squared and negative one. And of course, this time, we have zero π‘₯.

Let’s subtract one more time. π‘₯ squared minus π‘₯ squared is zero. Zero π‘₯ minus zero π‘₯ is zero. And zero minus negative one is one. So we can say that π‘₯ to the fourth power over π‘₯ squared minus one is π‘₯ squared plus one with the remainder of one. And we write that remainder as a fraction, as one over π‘₯ squared minus one.

Now, this is the bit, one over π‘₯ squared minus one, that we’re going to write using partial fraction form. Before we do though, we need to ensure that the denominator is fully factored. When we factor π‘₯ squared minus one, we get π‘₯ minus one times π‘₯ plus one. And now we need to write it in partial fraction form. Remember, essentially what we’re doing is we’re writing as the sum of two or more rational expressions. So we have 𝐴 over π‘₯ minus one plus 𝐡 over π‘₯ plus one, where 𝐴 and 𝐡 are constants.

Our job is to get this expression on the right-hand side to look like that on the left. And so we recall that, to add two algebraic fractions, we create a common denominator. To achieve this, we multiply the numerator and denominator of our first fraction by π‘₯ plus one and of our second fraction by π‘₯ minus one. And so that gives us on the right-hand side 𝐴 times π‘₯ plus one plus 𝐡 times π‘₯ minus one. And now their denominators are equal. So we have the numerators.

Now we know this is still equal to the fraction one over π‘₯ minus one times π‘₯ plus one. And since the denominators of these two fractions are equal, we know that their numerators must themselves be equal. So one must be equal to 𝐴 times π‘₯ plus one plus 𝐡 times π‘₯ minus one. And there are a couple of ways we can now find the values of 𝐴 and 𝐡. We could distribute our parentheses and then equate coefficients. But that can be quite a long-winded method. Alternatively, we substitute the zeros of π‘₯ minus one times π‘₯ plus one. That is, we let π‘₯ be equal to negative one and see what happens and let π‘₯ be equal to one.

When we let π‘₯ be equal to negative one, we get one equals 𝐴 times negative one plus one plus 𝐡 times negative one minus one because negative one plus one is zero. And so, by letting π‘₯ be equal to negative one, we have an equation purely in terms of 𝐡. We can see that one is equal to negative two 𝐡. And by dividing through by negative two, we find 𝐡 is equal to negative one-half.

We’ll now let π‘₯ be equal to one. This time, we get one equals 𝐴 times one plus one plus 𝐡 times one minus one. But of course, 𝐡 times one minus one is zero. So we get one equals two 𝐴. And when we divide through by two, we find 𝐴 to be equal to one-half. This means one over π‘₯ minus one times π‘₯ plus one can be written as a half over π‘₯ minus one plus negative a half over π‘₯ plus one. We can rewrite them further as one over two times π‘₯ minus one minus one over two times π‘₯ plus one. And we see we now need to integrate π‘₯ squared plus one plus these two partial fractions with respect to π‘₯. We could do this term by term.

When we integrate π‘₯ squared, we get π‘₯ cubed over three. And when we integrate one, we get π‘₯. We know that the integral of one over π‘₯ minus one is the natural log of the absolute value of π‘₯ minus one. So this third term integrates to a half times the natural log of π‘₯ minus one. And similarly, the fourth term integrates to negative one-half times the natural log of the absolute value of π‘₯ plus one. And then, we need that constant of integration π‘˜.

We can factor by one-half. And we get π‘₯ cubed over three plus π‘₯ plus a half times the natural log of the absolute value of π‘₯ minus one minus the natural log of the absolute value of π‘₯ plus one plus our constant π‘˜. Well we also know that the log of π‘Ž minus the log of 𝑏 is equal to the log of π‘Ž divided by 𝑏. So we get a half times the natural log of the absolute value of π‘₯ minus one over the absolute value of π‘₯ plus one. But the absolute value of π‘₯ minus one over the absolute value of π‘₯ plus one is equal to the absolute value of π‘₯ minus one over π‘₯ plus one.

And so we’ve evaluated the integral of π‘₯ to the fourth power over π‘₯ squared minus one with respect to π‘₯. It’s π‘₯ cubed over three plus π‘₯ plus a half times the natural log of the absolute value of π‘₯ minus one over π‘₯ plus one plus the constant of integration π‘˜.

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