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In this video, we’ll learn how to use partial fractions to evaluate integrals of rational functions with linear factors.
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It’s likely that you will have worked extensively with algebraic fractions, multiplying, dividing, adding, subtracting, and simplifying them.
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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.
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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.
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A single fraction with two distinct linear factors in the denominator can be split into two separate fractions with linear denominators.
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This is known as splitting it or decomposing it into partial fractions.
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And at this level, partial fractions are mainly used for binomial expansions and integration.
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To illustrate the method, let’s remind ourselves how we find the sum of two fractions with linear denominators.
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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.
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Here that’s three times 𝑥 plus five plus two times 𝑥 plus one over 𝑥 plus one times 𝑥 plus five.
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And distributing our parentheses, and we see that we’re left with five 𝑥 plus 17 over 𝑥 squared plus six 𝑥 plus five.
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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.
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But we now know that we can write this as three over 𝑥 plus one plus two over 𝑥 plus five.
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And then we recall the processes for integrating simple reciprocal functions.
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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.
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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.
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Well, that was all fine and well.
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But how do we figure out what our partial fractions are?
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Imagine we wanted to evaluate this indefinite integral.
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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.
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We’ll call their numerators 𝐴 and 𝐵.
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There are two ways to find the constants 𝐴 and 𝐵.
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Those are substitution and equating coefficients.
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We’ll also need to know what to do in the situation where we have an improper or a top heavy fraction.
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But let’s just begin with a simple example.
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Use partial fractions to evaluate the indefinite integral of 𝑥 plus four over 𝑥 plus six times 𝑥 minus one with respect to 𝑥.
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Remember, we need to rewrite this integrand using partial fractions.
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So we begin by reversing the process we would take when adding algebraic fractions.
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We write it as 𝐴 over 𝑥 plus six plus 𝐵 over 𝑥 minus one.
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And there are two ways for us to work out the constants 𝐴 and 𝐵.
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They’re substitution and equating coefficients.
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Let’s begin by looking at the substitution method.
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Let’s imagine we’re adding these fractions.
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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.
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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.
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Notice that the denominator of the fractions on both sides of our equation are equal.
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This means that, for the fractions themselves to be equal, their numerators must be.
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And we can say that 𝑥 plus four is equal to 𝐴 times 𝑥 minus one plus 𝐵 times 𝑥 plus six.
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Okay, so far, so good.
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Now we want to find a way to eliminate one of the constants from this equation.
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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.
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So let’s set 𝑥 be equal to one.
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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.
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And now we have an equation in 𝐵.
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So we can solve this by dividing both sides by seven.
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And we see that 𝐵 is equal to five-sevenths.
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Brilliant, so let’s repeat this process to help us establish the value of 𝐴.
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And you might wish to pause the video for a moment and think about what substitution would eliminate 𝐵 from this equation.
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If we let 𝑥 be equal to negative six, then the second term over here becomes 𝐵 times zero.
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So the 𝐵’s going to be eliminated.
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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.
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Now we have an equation in 𝐴.
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And dividing both sides by negative seven, we obtain 𝐴 to be equal to two-sevenths.
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And we’ve successfully decomposed into partial fractions.
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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.
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And we’re now of course able to integrate our expression with respect to 𝑥.
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Remembering that the integral of the sum of functions is the same as the sum of the integrals of those respective functions.
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And of course, we can take out any constant factors.
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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.
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Well, the integral of one over 𝑥 plus six is the natural log of the absolute value of 𝑥 plus six.
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And the integral of one over 𝑥 minus one is the natural log of the absolute value of 𝑥 minus one.
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And of course, since this is an indefinite integral, we must add that constant 𝑐.
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And we’re done.
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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.
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We are now going to consider how we could’ve got here using the method of equating coefficients.
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The starting process is the same.
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We need to get to this stage here.
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We want to distribute the parentheses on the right-hand side.
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And when we do, we see that 𝑥 plus four is equal to 𝐴𝑥 minus 𝐴 plus 𝐵𝑥 plus six 𝐵.
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Now this next step isn’t entirely necessary.
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But it can help us figure out what to do next.
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We collect together like terms.
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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.
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And now we have two families of terms, if you will.
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We have 𝑥 to the power of ones, and then we have these constants.
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And we can say that those are 𝑥 to the power of zeros.
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We want to equate coefficients for these terms.
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Let’s begin by equating the coefficients of 𝑥 to the power of zero, or just the constants.
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On the left-hand side, we have four.
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And on the right-hand side, we have six 𝐵 minus 𝐴.
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Next, we’ll equate the coefficients of 𝑥 to the power of one.
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The coefficient of 𝑥 on the left-hand side is one.
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And on the right-hand side, that’s 𝐴 plus 𝐵.
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So we now have a pair of simultaneous equations which we can begin to solve by first adding.
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Negative 𝐴 plus 𝐴 is zero.
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So we see that when we add our pair of simultaneous equations, we end up with five equals seven 𝐵.
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And solving this equation for 𝐵, we find that 𝐵 is equal to five-sevenths.
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We’re gonna then substitute this value of 𝐵 into either of our original equations.
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I’m going to choose this one here.
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So one is equal to 𝐴 plus five-sevenths.
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Then subtracting five-sevenths from both sides, we obtain 𝐴 to be equal to two-sevenths.
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And the rest of the process is exactly the same.
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We have our partial fractions, and we can integrate each of them.
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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.
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Both of these methods are equally as valid as one another.
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And there may even be times where you’ll need to combine the methods, which is also fine.
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We’ll now have a look at an example which involves three linear terms.
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Use partial fractions to evaluate the indefinite integral of one over 𝑡 cubed plus 𝑡 squared minus two 𝑡 with respect to 𝑡.
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When rewriting the integrand using partial fractions, we’ll be looking to reverse the process we would take when adding algebraic fractions.
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Before we can do this though, we need to decide what the denominator of these fractions might be.
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So let’s see if we can factor the denominator.
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It’s clearly a cubic, but there’s a common factor of 𝑡.
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So we’ll factor the denominator by 𝑡 to get 𝑡 times 𝑡 squared plus 𝑡 minus two.
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We can then factor this quadratic expression the usual way.
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𝑡 squared plus 𝑡 minus two is equal to 𝑡 minus one times 𝑡 plus two.
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We can now split this into partial fractions.
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These are 𝐴 over 𝑡 plus some other constant 𝐵 over 𝑡 minus one plus 𝐶 over 𝑡 plus two.
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Next, we’re going to add these three fractions.
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We’ll need to multiply the numerator and denominator of each of them by the denominator of the other two.
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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.
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And now we notice that the denominator of the fractions on both sides of this equation are equal.
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And for the fractions themselves to be equal, this means in turn that their numerators must be equal.
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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.
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We’re going to use the method of substitution to help us find the values for 𝐴, 𝐵, and 𝐶.
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Notice that if we let 𝑡 be equal to zero, we can instantly eliminate 𝐵 and 𝐶.
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So we substitute 𝑡 equals zero into our equation.
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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.
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Dividing both sides of this equation by negative two, and we obtain 𝐴 to be equal to negative one-half.
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Next, we spot that if we let 𝑡 be equal to one, we’re going to eliminate 𝐴 and 𝐶.
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Substituting 𝑡 equals one gives us one is equal to three 𝐵.
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And when we divide by three, we see that 𝐵 is equal to one-third.
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Finally, we’ll let 𝑡 be equal to negative two.
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This will eliminate 𝐴 and 𝐵 this time.
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This gives us one is equal to six 𝐶.
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And dividing through by six, we find 𝐶 to be equal to one-sixth.
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Let’s clear some space for the next step.
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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.
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We’re now ready to integrate with respect to 𝑡.
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The integral of one over 𝑡 is the natural log of the absolute value of 𝑡.
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So our first term integrates to negative a half times the natural log of 𝑡.
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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.
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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.
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We mustn’t forget since we’re dealing with an indefinite integral to add that constant of integration 𝐶.
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Now this method is fab.
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But it’s usually important to know that it can’t be used when there’s a repeated factor in the denominator.
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Let’s see what we’ll do in this case.
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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.
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Notice here that we have a repeated factor in the denominator of our fraction.
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A single fraction which has a repeated linear factor can be split into two or more separate fractions.
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But there is a special method for dealing with this linear factor.
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We list the repeated factor using increasing powers.
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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.
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Next, we’re going to add these fractions, remembering that the denominator is going to be 𝑡 times 𝑡 minus two all squared.
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So we multiply the numerator and denominator of our first fraction by 𝑡 minus two squared.
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For our second fraction, we multiply it by 𝑡 times 𝑡 minus two.
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And for our third fraction, we multiply that by simply 𝑡.
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And we see that the numerator is now 𝐴 times 𝑡 minus two all squared plus 𝐵𝑡 times 𝑡 minus two plus 𝐶𝑡.
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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.
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And we’ll use the method of substitution to find the values for 𝐴, 𝐵, and 𝐶.
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We’ll begin by letting 𝑡 be equal to two, with the aim of eliminating 𝐴 and 𝐵.
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On the left-hand side, we have three times two squared minus nine times two plus eight, which is two.
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And on the right-hand side, we have two 𝐶.
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Dividing through by two, and we find that 𝐶 is equal to one.
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Next, we’ll let 𝑡 be equal to zero.
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This time, that will eliminate 𝐵 and 𝐶.
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Substituting 𝑡 equals zero, we get eight equals four 𝐴.
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And then when we divide through by four, we find that 𝐴 is equal to two.
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The problem is, we’ve now run out of substitutions to make.
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So we’re going to use equating coefficients to find the value of 𝐵.
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So we distribute our parentheses.
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We’re going to equate the coefficients of 𝑡 squared.
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We have 𝐴 and 𝐵 on the right-hand side and three on the left.
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So we can say that three must be equal to 𝐴 plus 𝐵.
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We already established though that 𝐴 was equal to two.
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So we can substitute 𝐴 is equal to two and say that three is equal to two plus 𝐵.
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And subtracting two from both sides, we find that 𝐵 is equal to one.
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Let’s clear some space for the next step.
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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.
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We’re going to integrate this between the limits of one and 𝑥.
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The integral of two over 𝑡 is two times the natural log of 𝑡.
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The integral of one over 𝑡 minus two is the natural log of 𝑡 minus two.
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And then we can use the reverse chain rule on 𝑡 minus two to the power of negative two.
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And we find that the integral of that is negative one over 𝑡 minus two.
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We’re now going to substitute these limits in.
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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.
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This example is a nice example of how a mixture of substitution and equating coefficients can help us to achieve the solution.
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In our final example, we’ll consider what happens when we’re working with an improper fraction.
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Express 𝑥 cubed over 𝑥 squared plus two 𝑥 plus one in partial fraction form.
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Notice that the degree of our numerator is higher than the degree of the denominator.
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On the numerator, we have 𝑥 cubed.
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And on the denominator, we only go as far as 𝑥 squared.
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This means that we know that we have an improper fraction.
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And we’re going to need to perform polynomial long division.
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You might wish to pause the video and divide 𝑥 cubed by 𝑥 squared plus two 𝑥 plus one yourself.
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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.
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We now have a mixed fraction.
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Factoring the denominator of the remainder part, and we notice we then have a repeated root.
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So when writing in impartial fraction form, we write it as 𝐴 over 𝑥 plus one plus 𝐵 over 𝑥 plus one squared.
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When we add the fractions, we multiply the first one by 𝑥 plus one.
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So we have 𝐴 times 𝑥 plus one.
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And we don’t need to do anything with the second.
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So we find that, in the numerators, we have three 𝑥 plus two equals 𝐴 times 𝑥 plus one plus 𝐵.
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We’ll begin by equating coefficients of 𝑥 to the power of one.
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On the left-hand side, that’s three.
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And on the right-hand side, that’s 𝐴.
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So we see that 𝐴 is equal to three.
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We then equate coefficients of 𝑥 to the power of zero or constants.
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And we find that two is equal to 𝐴 plus 𝐵.
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But we know that 𝐴 is equal to three.
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So two is equal to three plus 𝐵.
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We then subtract three from both sides, and we find 𝐵 to be equal to negative one.
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And we have expressed our fraction in partial fraction form.
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It’s 𝑥 minus two plus three over 𝑥 plus one minus one over 𝑥 plus one squared.
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And if we’re so required, we could now integrate it.
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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.
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This is called splitting into partial fractions.
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And we saw that this can make the integration process considerably easier.
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We also saw that this method can be applied when there are more than two distinct linear factors.
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But that a single fraction with a repeated linear factor needs a special rule whereby we list the repeated power in ascending powers.
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Finally, we learned that if the fraction is improper, its numerator has a degree equal to or larger than the denominator.
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And we need to convert it to a mixed fraction by using polynomial long division before it can be expressed as partial fractions.