Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ (2π‘₯Β³ βˆ’ 4π‘₯Β² βˆ’ π‘₯ βˆ’ 3)/(π‘₯Β² βˆ’ 2π‘₯ βˆ’ 3) dπ‘₯.

08:26

Video Transcript

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‘₯.

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. So the numerator has a larger π‘₯ term than the denominator. And in order to use partial fractions on this fraction, we need the denominator to have the larger power of π‘₯.

So first, we need to do some division and divide the numerator by the denominator. In order to do this division, will use algebraic long division. 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. And so, essentially, we’re doing two π‘₯ cubed divided by π‘₯ squared. And clearly, this is equal to two π‘₯. So we can write two π‘₯ up here.

And now we subtract two π‘₯ timesed by the polynomial we’re dividing by from the polynomial that we’re dividing. So what we are subtracting here is two π‘₯ lots of π‘₯ squared minus two π‘₯ minus three. This is equal to two π‘₯ cubed minus four π‘₯ squared minus six π‘₯. So let’s subtract that from the polynomial which we’re dividing here. And so we get that the two π‘₯ cubed cancels with the two π‘₯ cubed. Then we have minus minus four π‘₯, which is simply plus four π‘₯. So that cancels with the negative four π‘₯ in the polynomial. And then we have minus minus six π‘₯. So we are adding six π‘₯ here. So minus π‘₯ plus six π‘₯ leaves us with five π‘₯, and minus three remains unchanged.

And now we want to see how many times π‘₯ squared will go into five π‘₯. However, since the π‘₯ squared term is larger than the five π‘₯ term, we get zero here. And so this is as far as we can go with this division.

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. However, we mustn’t forget the five π‘₯ minus three. And this is called a remainder here. And this gets added on to the right-hand side of the equation. 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. And now we can substitute this back into the numerator of the fraction in our original integral. And so we obtain this.

Now we can split the fraction on the right into two. 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. And now we can split the fraction up on the right into partial fractions.

We will start by factorising the denominator. 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. Next, we will write this in partial fraction forms with the numerators as constants which we need to find.

So this is equal to 𝐴 over π‘₯ minus three plus 𝐡 over π‘₯ plus one. And next, we will multiply through by the denominator of this fraction, leaving us with five π‘₯ minus three is equal to 𝐴 lots of π‘₯ plus one plus 𝐡 lots of π‘₯ minus three. And now we expand the brackets on the right to get 𝐴π‘₯ plus 𝐴 plus 𝐡π‘₯ minus three 𝐡. And now we can group together the π‘₯ terms and the constants, which leaves us with 𝐴 plus 𝐡 lots of π‘₯ plus 𝐴 minus three 𝐡.

Next, we will compare the coefficients on the left and right of the equation. So we see that, on the left, we have five π‘₯. And on the right, we have 𝐴 plus 𝐡π‘₯. This tells us that 𝐴 plus 𝐡 equals five. And comparing the constant term, we see that we have minus three on the left and 𝐴 minus three 𝐡 on the right, which tells us that 𝐴 minus three 𝐡 is equal to minus three.

Now we can rearrange the equation on the right, adding three 𝐡 to both sides, which leaves us with 𝐴 is equal to three 𝐡 minus three. And now we can substitute this in for 𝐴 in the other equation, which tells us that three 𝐡 minus three plus 𝐡 is equal to five. And we add three to both sides to give us four 𝐡 is equal to eight. And now we divide by four. And we find that 𝐡 is equal to two.

And now we can substitute this value of 𝐡 equals two into the equation 𝐴 plus 𝐡 equals five. And this tells us that 𝐴 plus two is equal to five. Subtracting three from both sides tells us that 𝐴 is equal to three. And so now we have found our values for 𝐴 and 𝐡.

Now we substitute these values for 𝐴 and 𝐡 back into the equation. 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. 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. 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‘₯. And this is an integral which we can calculate.

We will use the fact that the integral of one over π‘₯ plus 𝐴 with respect to π‘₯, where 𝐴 is a constant, is equal to the natural logarithm of the absolute value of π‘₯ plus 𝐴. 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. So we need to divide by two.

For the next two terms, we’re using the rule that we have just stated. 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. 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 we’re integrating here. And we can do this since these are both just constants.

Now for the final term here, we mustn’t forget that this is an indefinite integral. So we need to add on the constant of integration 𝐢. 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 𝐢.

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