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Use partial fractions to evaluate the indefinite integral of two π₯ cubed minus four π₯ squared minus π₯ minus three all over π₯ squared minus two π₯ minus three with respect to π₯.
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Firstly, we notice that the order of the largest π₯ term in the numerator is three and the order of the largest π₯ term in the denominator is two.
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So the numerator has a larger π₯ term than the denominator.
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And in order to use partial fractions on this fraction, we need the denominator to have the larger power of π₯.
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So first, we need to do some division and divide the numerator by the denominator.
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In order to do this division, will use algebraic long division.
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The first step we need to do here is work out how many times the largest terms in this polynomial goes into the largest term in this polynomial.
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And so, essentially, weβre doing two π₯ cubed divided by π₯ squared.
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And clearly, this is equal to two π₯.
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So we can write two π₯ up here.
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And now we subtract two π₯ timesed by the polynomial weβre dividing by from the polynomial that weβre dividing.
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So what we are subtracting here is two π₯ lots of π₯ squared minus two π₯ minus three.
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This is equal to two π₯ cubed minus four π₯ squared minus six π₯.
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So letβs subtract that from the polynomial which weβre dividing here.
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And so we get that the two π₯ cubed cancels with the two π₯ cubed.
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Then we have minus minus four π₯, which is simply plus four π₯.
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So that cancels with the negative four π₯ in the polynomial.
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And then we have minus minus six π₯.
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So we are adding six π₯ here.
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So minus π₯ plus six π₯ leaves us with five π₯, and minus three remains unchanged.
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And now we want to see how many times π₯ squared will go into five π₯.
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However, since the π₯ squared term is larger than the five π₯ term, we get zero here.
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And so this is as far as we can go with this division.
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What this has told us is that there are two π₯ lots of π₯ squared minus two π₯ minus three in two π₯ cubed minus four π₯ squared minus π₯ minus three.
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However, we mustnβt forget the five π₯ minus three.
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And this is called a remainder here.
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And this gets added on to the right-hand side of the equation.
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So this gives us that two π₯ cubed minus four π₯ squared minus π₯ minus three is equal to two π₯ lots of π₯ squared minus two π₯ minus three plus five π₯ minus three.
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And now we can substitute this back into the numerator of the fraction in our original integral.
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And so we obtain this.
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Now we can split the fraction on the right into two.
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And so the left-hand fraction here now simplifies down to give us two π₯ plus five π₯ minus three over π₯ squared minus two π₯ minus three.
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And now we can split the fraction up on the right into partial fractions.
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We will start by factorising the denominator.
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And so we get that five π₯ minus three over π₯ squared minus two π₯ minus three is also equal to five π₯ minus three over π₯ minus three times π₯ plus one.
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Next, we will write this in partial fraction forms with the numerators as constants which we need to find.
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So this is equal to π΄ over π₯ minus three plus π΅ over π₯ plus one.
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And next, we will multiply through by the denominator of this fraction, leaving us with five π₯ minus three is equal to π΄ lots of π₯ plus one plus π΅ lots of π₯ minus three.
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And now we expand the brackets on the right to get π΄π₯ plus π΄ plus π΅π₯ minus three π΅.
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And now we can group together the π₯ terms and the constants, which leaves us with π΄ plus π΅ lots of π₯ plus π΄ minus three π΅.
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Next, we will compare the coefficients on the left and right of the equation.
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So we see that, on the left, we have five π₯.
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And on the right, we have π΄ plus π΅π₯.
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This tells us that π΄ plus π΅ equals five.
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And comparing the constant term, we see that we have minus three on the left and π΄ minus three π΅ on the right, which tells us that π΄ minus three π΅ is equal to minus three.
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Now we can rearrange the equation on the right, adding three π΅ to both sides, which leaves us with π΄ is equal to three π΅ minus three.
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And now we can substitute this in for π΄ in the other equation, which tells us that three π΅ minus three plus π΅ is equal to five.
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And we add three to both sides to give us four π΅ is equal to eight.
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And now we divide by four.
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And we find that π΅ is equal to two.
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And now we can substitute this value of π΅ equals two into the equation π΄ plus π΅ equals five.
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And this tells us that π΄ plus two is equal to five.
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Subtracting three from both sides tells us that π΄ is equal to three.
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And so now we have found our values for π΄ and π΅.
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Now we substitute these values for π΄ and π΅ back into the equation.
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And we get that five π₯ minus three over π₯ minus three times π₯ minus one is equal to three over π₯ minus three plus two over π₯ minus one.
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What we have found here is that two π₯ cubed minus four π₯ squared minus π₯ minus three all over π₯ squared minus two π₯ minus three is equal to two π₯ plus three over π₯ minus three plus two over π₯ plus one.
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And so we can substitute this fraction into our integral, leaving us with the indefinite integral of two π₯ plus three over π₯ minus three plus two over π₯ plus one with respect to π₯.
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And this is an integral which we can calculate.
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We will use the fact that the integral of one over π₯ plus π΄ with respect to π₯, where π΄ is a constant, is equal to the natural logarithm of the absolute value of π₯ plus π΄.
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And now when we integrate this, which we can do term by term, for the first term, two π₯, we increase the power by one β so we get two π₯ squared β and then divide by the new power.
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So we need to divide by two.
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For the next two terms, weβre using the rule that we have just stated.
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And so we get plus three times the natural logarithm of the absolute value of π₯ minus three plus two times the natural logarithm of the absolute value of π₯ plus one.
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Now this three and this two here have come from the fact that we have a three in the numerator and two in the numerator of the two fractions, which weβre integrating here.
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And we can do this since these are both just constants.
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Now for the final term here, we mustnβt forget that this is an indefinite integral.
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So we need to add on the constant of integration πΆ.
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We can now simplify this to obtain a final answer that the indefinite integral of two π₯ cubed minus four π₯ squared minus π₯ minus three all over π₯ squared minus two π₯ minus three with respect to π₯ is equal to π₯ squared plus three times the natural logarithm of the absolute value of π₯ minus three plus two times the natural logarithm of the absolute value of π₯ plus one plus πΆ.