Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫((6π‘₯ + 7)/(π‘₯ + 2)Β²) dπ‘₯.

05:49

Video Transcript

Use partial fractions to evaluate the integral of six π‘₯ plus seven over π‘₯ plus two all squared with respect to π‘₯.

We are told to use partial fractions to evaluate this integral. So, we’ll begin by splitting the fraction into its parts. And we can only do this once the denominator is in factorised form. Our denominator is already in factorised form. So, we can say that our expression is equal to some number over π‘₯ plus two plus some other number over π‘₯ plus two all squared.

Remember, since the fact that π‘₯ plus two actually appears twice, we account for this by forming fractions containing increasing powers of π‘₯ plus two, as shown. What we do next is multiply both sides of this identity by the denominator of our fraction on the left-hand side. This has the effect of cancelling out each of the denominators. And on the left-hand side, we’re left with six π‘₯ plus seven.

On the right, it’s a little trickier. When we multiply by π‘₯ plus two all squared, one factor of π‘₯ plus two cancels in this first term. And we’re just left with 𝐴 multiplied by π‘₯ plus two. And with the second partial fraction, π‘₯ plus two all squared cancels through on both the numerator and the denominator. And so, we’re simply left with 𝐡. And we can now say that six π‘₯ plus seven is equal to 𝐴 multiplied by π‘₯ plus two plus 𝐡. And in fact, this is an identity. It’s true for all values of π‘₯.

Next, there are several methods for determining the coefficient of each term. The most efficient way here in my opinion is to equate coefficients. We begin by expanding this bracket. 𝐴 multiplied by π‘₯ is 𝐴π‘₯. And 𝐴 multiplied by two is two 𝐴. So, we see that six π‘₯ plus seven is always equal to 𝐴π‘₯ plus two 𝐴 plus 𝐡. And this method is called equating coefficients because we start to look at the coefficients of our various powers of π‘₯.

We can see in this example we have π‘₯ to the power of one and we have π‘₯ to the power of zero. We’ll begin by equating coefficients of π‘₯ to the power of one. On the left-hand side, that’s six. And on the right-hand side, that’s 𝐴. So, we can see that 𝐴 must be equal to six. Then, we’ll equate the coefficient of π‘₯ to the power of zero. These are actually just the constants on both sides of our identity. On the left, that’s seven. And on the right, that’s two 𝐴 plus 𝐡. So, two 𝐴 plus 𝐡 must be equal to seven.

In fact though, we’ve just seen that 𝐴 is equal to six. So, we can say that seven is equal to two multiplied by six plus 𝐡, or seven is equal to 12 plus 𝐡. And if we subtract 12 from both sides, we see that 𝐡 is equal to negative five. This means we can now substitute our values of 𝐴 and 𝐡 into our earlier equation. And we see that six π‘₯ plus seven over π‘₯ plus two all squared must be equal to six over π‘₯ plus two plus negative five π‘₯ plus two squared.

And all that’s left is to integrate this with respect to π‘₯. It can be helpful to split this integral up. And in fact, I like to take the constant factor outside each integral sign. So, we have six lots of the integral of one over π‘₯ plus two with respect to π‘₯ minus five lots of the integral of one over π‘₯ plus two all squared with respect to π‘₯.

And here, we recall the fact that the integral of 𝑓 dash of π‘₯ over 𝑓 of π‘₯ with respect to π‘₯ is equal to ln of 𝑓 of π‘₯ plus the constant of integration 𝑐. In the expression one over π‘₯ plus two, π‘₯ plus two is our function of π‘₯. And if we were to differentiate that function, we’d simply get one. And so, this is an example of where we can use this general rule. And we can say that the integral of one over π‘₯ plus two with respect to π‘₯ is ln of π‘₯ plus two plus that constant of integration.

And, of course, we took a constant six outside the integral sign, so we have to remember to multiply that all by six. Now notice, we’ve used to 𝑐 one as our constant of integration because we’re going to be integrating a couple of different expressions here.

Next, we need to integrate one over π‘₯ plus two all squared. Here, we’re going to integrate using substitution. And we’re going to let 𝑒 be equal to π‘₯ plus two. d𝑒 dπ‘₯ here is, therefore, equal to one. And we can say that d𝑒 must, therefore, be equal to dπ‘₯. And then, we replace π‘₯ plus two with 𝑒 and dπ‘₯ with d𝑒. And now we’re integrating one over 𝑒 squared with respect to 𝑒. In fact, it’s often easier to think about this as the integral of 𝑒 to the power of negative two with respect to 𝑒.

When we integrate, we raise the power by one and then divide by that number. So, negative two plus one is negative one. And we have 𝑒 to the power of negative one divided by negative one, which we can say is negative 𝑒 to the negative one plus a second constant of integration. And then, we can replace this with that integral. And we’re left with six of ln of π‘₯ plus two plus our first constant minus five multiplied by negative π‘₯ plus two to the power of negative one plus our second constant of integration.

And what’s going to happen with these constants of integration is one is going to be multiplied by six. The other is going to be multiplied by negative five. And we’re going to add them together to form a new constant of integration. Let’s call that π‘˜.

And distributing the other part of our first bracket, we get six multiplied by ln of π‘₯ plus two. And negative five multiplied by negative π‘₯ plus two to the power of negative one is positive five multiplied by π‘₯ plus two to the power of negative one. And there we go. We’ve used partial fractions to evaluate our integral. It’s six ln of π‘₯ plus two plus five multiplied by π‘₯ plus two to the power of negative one plus π‘˜.

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