Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ (π‘₯Β² + 4π‘₯ + 1)/((π‘₯ βˆ’ 1)(π‘₯ + 1)(π‘₯ + 3)) dπ‘₯.

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Video Transcript

Use partial fractions to evaluate the integral of π‘₯ squared plus four π‘₯ plus one all over π‘₯ minus one multiplied by π‘₯ plus one multiplied by π‘₯ plus three with respect to π‘₯.

We’re 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 when the denominator is in factorised form. The denominator of our fraction is already in factorised form. So, we can say that it’s equal to some number over π‘₯ minus one plus some other number over π‘₯ plus one plus some other number over π‘₯ plus three.

What we do in π‘₯ is multiply it by the denominator of our fraction. This will have the effect of cancelling out each of the denominators. On the left-hand side, we are left with π‘₯ squared plus four π‘₯ plus one. On the right, it’s a little trickier. When we multiply this first term by the denominator, the π‘₯ minus ones cancel. And that means we’re left with 𝐴 multiplied by π‘₯ plus one multiplied by π‘₯ plus three.

With our second fraction, π‘₯ plus one cancels and we’re left with 𝐡 multiplied by π‘₯ minus one multiplied by π‘₯ plus three. And with our last fraction, the π‘₯ plus three cancels. So, we’re left with 𝐢 multiplied by π‘₯ minus one multiplied by π‘₯ plus one.

Now at this stage, we can do one of two things. We could expand all of the brackets and compare coefficients. But that can be quite a long process. Instead, we recognise that this is an identity. And that means it’s true for every value of π‘₯. So, we can substitute any values of π‘₯ into this identity. In fact, we choose to substitute the roots of the denominator into the identity and use that information to solve for 𝐴, 𝐡, and 𝐢. Let’s see what this looks like.

One of the roots is π‘₯ is equal to one. Substituting one into the left-hand side of our identity gives us one squared plus four multiplied by one plus one, which is six. And on the right-hand side, we get 𝐴 multiplied by two multiplied by four, which is eight 𝐴. And in fact, the other two terms disappear because we have π‘₯ minus one, which we now know is one minus one, which gives us zero. So, we’re left with six is equal to eight 𝐴. And we solve this equation for 𝐴 by dividing through by eight. So, 𝐴 is equal to six-eighths, which, of course, simplifies to three-quarters. We’ll repeat this with the next root.

Another root of this equation is π‘₯ is equal to negative one. When we substitute negative one into the left-hand side of our identity, we get negative one squared plus four multiplied by negative one plus one, which is negative two. This time the coefficient of 𝐴 and 𝐢 is zero. So, we have on the right-hand side 𝐡 is equal to negative one minus one, which is negative two, multiplied by negative one plus three, which is two. So, negative two is equal to negative four 𝐡. And dividing through by negative four, we see that 𝐡 is equal to one-half.

Finally, we’re interested in the third root. And that’s when π‘₯ is equal to negative three. This time on the left-hand side, we get negative three squared plus four multiplied by negative three plus one, which is negative two. And this time, the first and second terms, the coefficient of 𝐴 and 𝐡 are zero.

So, we have 𝐢 multiplied by π‘₯ minus one, which becomes negative three minus one, which is negative four, multiplied by π‘₯ plus one negative three plus one, which is negative two. And we can see that negative two is equal to eight 𝐢. This time to solve for 𝐢, we divide through by eight. And we’re left with 𝐢 is equal to negative one-quarter.

We now have enough information to write our integral in partial fraction form. We substitute 𝐴, 𝐡, and 𝐢 into our earlier equation. And we see that our integral is equal to the integral of three over four lots of π‘₯ minus one plus one over two lots of π‘₯ plus one minus one over four lots of π‘₯ plus three. All that’s left is to integrate this.

It can help to split this integral up and take the constant factor outside each integral sign. So, we get three-quarters of the integral of one over π‘₯ minus one with respect to π‘₯ and so on. And then we recall this fact. The integral of 𝑓 dash π‘₯ over 𝑓 of π‘₯ with respect to π‘₯ is equal to ln of 𝑓 of π‘₯ plus 𝑐. And the integral of one over π‘₯ minus one is ln of π‘₯ minus one plus 𝑐. And all of that’s multiplied by three-quarters.

In fact, we’re going to be dealing with a few constants of integration here, so let’s call that 𝑐 one. Integrating one over π‘₯ plus one with respect to π‘₯, and we get ln of π‘₯ plus one plus our constant of integration. And we multiply all of that by one-half. And finally, we integrate one over π‘₯ plus three with respect to π‘₯ and we get ln of π‘₯ plus three plus another constant of integration. And all of that, this time, is multiplied by negative one-quarter.

Let’s simplify this somewhat. We can combine each of these constants of integration. Let’s call that 𝐾. So, we’re left with three-quarters of ln of π‘₯ minus one plus one-half of ln of π‘₯ plus one minus one-quarter of ln of π‘₯ plus three plus our new constant of integration 𝐾. And we aren’t required to simplify this using the laws of logs, so we’re done.

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