WEBVTT
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Use partial fractions to evaluate the indefinite integral of π₯ to the fourth power of π₯ squared minus one with respect to π₯.
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Now, we are actually told how weβre going to evaluate this integral.
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We need to rewrite our integrand β thatβs π₯ to the fourth power over π₯ squared minus one β in partial fraction form.
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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.
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And that means π₯ to the fourth power over π₯ squared minus one is a top-heavy or an improper fraction.
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And so, before we can write in partial fraction form, weβre going to use polynomial long division to divide π₯ to the fourth power by π₯ squared minus one.
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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.
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It just makes the whole process of polynomial long division a little bit easier.
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So the first thing we do is divide π₯ to the fourth power by π₯ squared.
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Thatβs π₯ squared.
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We then multiply π₯ squared by π₯ squared and π₯ squared by negative one.
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That gives us π₯ to the fourth power minus π₯ squared.
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And of course, in this term, we have zero π₯ cubed.
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We now subtract these three terms.
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π₯ to the fourth power minus π₯ to the fourth power is zero.
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Zero π₯ cubed minus zero π₯ cubed is zero.
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And zero π₯ squared minus negative π₯ squared is π₯ squared.
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We bring down the remaining terms.
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And now, we divide π₯ squared by π₯ squared, which is simply one.
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This time, we multiply one by each term in our divisor.
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Thatβs π₯ squared and negative one.
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And of course, this time, we have zero π₯.
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Letβs subtract one more time.
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π₯ squared minus π₯ squared is zero.
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Zero π₯ minus zero π₯ is zero.
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And zero minus negative one is one.
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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.
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And we write that remainder as a fraction, as one over π₯ squared minus one.
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Now, this is the bit, one over π₯ squared minus one, that weβre going to write using partial fraction form.
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Before we do though, we need to ensure that the denominator is fully factored.
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When we factor π₯ squared minus one, we get π₯ minus one times π₯ plus one.
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And now we need to write it in partial fraction form.
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Remember, essentially what weβre doing is weβre writing as the sum of two or more rational expressions.
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So we have π΄ over π₯ minus one plus π΅ over π₯ plus one, where π΄ and π΅ are constants.
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Our job is to get this expression on the right-hand side to look like that on the left.
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And so we recall that, to add two algebraic fractions, we create a common denominator.
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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.
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And so that gives us on the right-hand side π΄ times π₯ plus one plus π΅ times π₯ minus one.
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And now their denominators are equal.
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So we have the numerators.
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Now we know this is still equal to the fraction one over π₯ minus one times π₯ plus one.
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And since the denominators of these two fractions are equal, we know that their numerators must themselves be equal.
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So one must be equal to π΄ times π₯ plus one plus π΅ times π₯ minus one.
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And there are a couple of ways we can now find the values of π΄ and π΅.
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We could distribute our parentheses and then equate coefficients.
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But that can be quite a long-winded method.
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Alternatively, we substitute the zeros of π₯ minus one times π₯ plus one.
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That is, we let π₯ be equal to negative one and see what happens and let π₯ be equal to one.
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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.
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And so, by letting π₯ be equal to negative one, we have an equation purely in terms of π΅.
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We can see that one is equal to negative two π΅.
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And by dividing through by negative two, we find π΅ is equal to negative one-half.
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Weβll now let π₯ be equal to one.
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This time, we get one equals π΄ times one plus one plus π΅ times one minus one.
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But of course, π΅ times one minus one is zero.
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So we get one equals two π΄.
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And when we divide through by two, we find π΄ to be equal to one-half.
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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.
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We can rewrite them further as one over two times π₯ minus one minus one over two times π₯ plus one.
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And we see we now need to integrate π₯ squared plus one plus these two partial fractions with respect to π₯.
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We could do this term by term.
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When we integrate π₯ squared, we get π₯ cubed over three.
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And when we integrate one, we get π₯.
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We know that the integral of one over π₯ minus one is the natural log of the absolute value of π₯ minus one.
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So this third term integrates to a half times the natural log of π₯ minus one.
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And similarly, the fourth term integrates to negative one-half times the natural log of the absolute value of π₯ plus one.
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And then, we need that constant of integration π.
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We can factor by one-half.
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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.
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Well we also know that the log of π minus the log of π is equal to the log of π divided by π.
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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.
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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.
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And so weβve evaluated the integral of π₯ to the fourth power over π₯ squared minus one with respect to π₯.
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Itβs 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.