# Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate β« π₯β΄/(π₯Β² β 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 weβre going to evaluate this integral. We need to rewrite our integrand β thatβ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, weβ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. Thatβ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. Thatβs π₯ squared and negative one. And of course, this time, we have zero π₯.

Letβ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 weβ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 weβre doing is weβre writing as the sum of two or more rational expressions. So we have π΄ over π₯ minus one plus π΅ over π₯ plus one, where π΄ and π΅ 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 π΄ times π₯ plus one plus π΅ 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 π΄ times π₯ plus one plus π΅ times π₯ minus one. And there are a couple of ways we can now find the values of π΄ and π΅. 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 π΄ times negative one plus one plus π΅ 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 π΅. We can see that one is equal to negative two π΅. And by dividing through by negative two, we find π΅ is equal to negative one-half.

Weβll now let π₯ be equal to one. This time, we get one equals π΄ times one plus one plus π΅ times one minus one. But of course, π΅ times one minus one is zero. So we get one equals two π΄. And when we divide through by two, we find π΄ 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 minus the log of π is equal to the log of π divided by π. 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 weβve evaluated the integral of π₯ to the fourth power over π₯ squared minus one with respect to π₯. Itβ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.