Video: Integration of Rational Functions by Partial Fractions

Video Transcript

Use partial fractions to evaluate the indefinite integral of eight 𝑥 squared plus eight 𝑥 plus two over four 𝑥 squared plus one all squared.

Now, we are actually told how we’re going to evaluate this integral. We’re going to write the fraction — the integrand — in partial faction form. But we’re going to need to be really careful. And there’s two reasons for this. Firstly, one of the factors in our denominator is a nonfactorable quadratic and, secondly, that quadratic is being squared. And so, to deal with this, there are two things we need to do. We separate our fraction into the sum of two rational functions. Because the denominators of both of these rational functions are made up of these quadratics, the numerators must be of the form 𝐴𝑥 plus 𝐵 and 𝐶𝑥 plus 𝐷, where 𝐴, 𝐵, 𝐶, and 𝐷 are constants. And because that quadratic expression is being squared, the numerators are four 𝑥 squared plus one and four 𝑥 squared plus one squared.

Now, we need to make the expression on the right look like that on the left. And to do so, we do something with this first fraction here. We multiply both the numerator and denominator of this fraction by four 𝑥 squared plus one. In doing so, this creates a common denominator. And then, we’ll simply be able to add the numerators. We get 𝐴𝑥 plus 𝐵 times four 𝑥 squared plus one plus 𝐶𝑥 plus 𝐷 all over four 𝑥 squared plus one squared. Now, this is, of course, equal to the earlier fraction. And we can see that the denominators of each of our fractions are equal, which means, in turn, their numerators must also be equal. That is to say, eight 𝑥 squared plus eight 𝑥 plus two is equal to 𝐴𝑥 plus 𝐵 times four 𝑥 squared plus one plus 𝐶𝑥 plus 𝐷.

Our job now is to find the values of the constants 𝐴, 𝐵, 𝐶, and 𝐷. Now, usually, we’d look to substitute the zeros of four 𝑥 squared plus one in. But we don’t have any here. And so, what we’re going to do is distribute the parentheses on the right-hand side and equate coefficients. Distributing the parentheses and we get four 𝐴𝑥 cubed plus 𝐴𝑥 plus four 𝐵𝑥 squared plus 𝐵 plus 𝐶𝑥 plus 𝐷. We’re going to begin then by equating coefficients of 𝑥 cubed. On the left-hand side, there are no 𝑥 cubed terms. So, the coefficient of 𝑥 cubed is zero. And on the right, the only coefficient we have is four 𝐴. Dividing through by four then and we find that 𝐴 itself is equal to zero.

Let’s repeat this process for 𝑥 squared. This time, on the left, we have eight. And on the right, it’s four 𝐵. Dividing both sides of this equation by four and we find that 𝐵 must be equal to two. Now, we’ll create coefficients of 𝑥 or 𝑥 to the power of one. On the left, that’s eight, and then on the right, we now have two terms. We have 𝐴 and 𝐶. But of course, we said that 𝐴 is equal to zero. And that means that 𝐶 must be equal to eight.

Finally, we equate the coefficients of 𝑥 to the power of zero. But of course, those are just the constant terms. On the left, that’s two. And on the right, we have 𝐵 and 𝐷. Remember, though, we said that 𝐵 is equal to two. And so, if we subtract two from both sides of this equation, we find 𝐷 is equal to zero. And so, going back to our partial fraction form, we see that we can write our fraction as zero 𝑥 plus two over four 𝑥 squared plus one plus eight 𝑥 plus zero over four 𝑥 squared plus one all squared.

And we’re now ready to integrate. Let’s clear some space. We can see that we’re integrating two over four 𝑥 squared plus one plus eight 𝑥 over four 𝑥 squared plus one all squared with respect to 𝑥. Let’s do this term by term. We’re first going to look at the expression two over four 𝑥 squared plus one. Here, it’s really useful to recall that the derivative of one over one plus 𝑥 squared is equal to the inverse tan of 𝑥. We can take out constant factors. So, we’ll take out that factor of two. And then, we’re going to write four 𝑥 squared as two 𝑥 all squared. And then, this means when we integrate one over one plus two 𝑥 all squared, we get the inverse tan of two 𝑥. So, the integral of two times this is two times the inverse tan of two 𝑥.

And what about the integral of eight 𝑥 over four 𝑥 squared plus one all squared? Well, if we spot that the numerator is a scalar multiple of the derivative of the inner function in our denominator, that means we can use integration by substitution. Now, we let 𝑢 be equal to that in a function, four 𝑥 squared plus one. Then, d𝑢 by d𝑥, the derivative of this expression with respect to 𝑥, is eight 𝑥. Now, d𝑢 by d𝑥 isn’t a fraction, but we do treat it a little like one. And we can write this as d𝑢 equals eight 𝑥 d𝑥. Then, we replace eight 𝑥 d𝑥 with d𝑢 and of course four 𝑥 squared plus one with 𝑢. So, we need to integrate one over 𝑢 squared with respect to 𝑢 or 𝑢 to the power of negative two. To integrate this, we add one to the power and divide by that new value. So, it’s 𝑢 to the power of negative one over negative one, which is negative one over 𝑢.

Our final step with this substitution is to replace 𝑢 with four 𝑥 squared plus one. And so, we see that the integral of eight 𝑥 over four 𝑥 squared plus one all squared is negative one over four 𝑥 squared plus one. This is an indefinite integral. So, we add our constant of integration. And we see that the integral of eight 𝑥 squared plus eight 𝑥 plus two all over four 𝑥 squared plus one all squared is two times the inverse tan of two 𝑥 minus one over four 𝑥 squared plus one plus our constant of integration 𝑘.