### Video Transcript

Use partial fractions to evaluate the integral of 16π₯ cubed over four π₯ squared minus four π₯ plus one with respect to π₯.

Now, we are actually told how we need to solve this problem. We need to write 16π₯ cubed over four π₯ squared minus four π₯ plus one as partial fractions. Before we can though, we need to notice that the highest power in the numerator is greater than the highest power of π₯ in the denominator. And so, this is a top-heavy fraction. Weβre going to need to use polynomial long division before we can actually use partial fraction decomposition. And so, weβre going to divide 16π₯ cubed by four π₯ squared minus four π₯ plus one. Now, letβs write 16π₯ cubed as 16π₯ cubed plus zero π₯ squared plus zero π₯ plus zero. This just makes the following steps a little easier.

The first thing we do then is divide 16π₯ cubed by four π₯ squared. Thatβs four π₯. Then, we individually multiply four π₯ by four π₯ squared, negative four π₯, and one. That gives us 16π₯ cubed minus 16π₯ squared plus four π₯. The next thing we do is we subtract each of these terms. 16π₯ cubed minus 16π₯ cubed is zero. Zero π₯ squared minus negative 16π₯ squared is 16π₯ squared. And zero π₯ minus four π₯ is negative four π₯. Letβs bring down this zero. Next, we divide 16π₯ squared by four π₯ squared, which is simply four. And then, we multiply this four by each term in the denominator. That gives us 16π₯ squared minus 16π₯ plus four. And then we subtract one last time. And weβre left with 12π₯ minus four.

Now, what this actually means is we can write 16π₯ cubed over four π₯ squared minus four π₯ plus one as four π₯ plus four plus 12π₯ minus four all over four π₯ squared minus four π₯ plus one. Now, this part here is what weβre actually going to write in partial fraction form. Before we can though, we need to ensure that the denominator is fully factored. And when we factor four π₯ squared minus four π₯ plus one, we get two π₯ minus one all squared. And now, we can write our fraction as the sum of two rational functions. Itβs π΄ over two π₯ minus one plus two π΅ over two π₯ minus one all squared, where π΄ and π΅ are constants.

Weβre going to add the fractions on the right-hand side by creating a common denominator of two π₯ minus one squared. To achieve this, we multiply the numerator and denominator of our first fraction by two π₯ minus one. And so, we get π΄ times two π₯ minus one over two π₯ minus one squared plus π΅ over two π₯ minus one squared. And since the denominators are equal, this is π΄ times two π₯ minus one plus π΅ over two π₯ minus one squared. Now of course, these two fractions are equal. And we can see now that their denominators are equal. So, this must mean their numerators themselves are equal.

In other words, 12π₯ minus four must be equal to π΄ times two π₯ minus one plus π΅. And we have a couple of ways that we can solve this. We can substitute in roots or we can distribute the parentheses and acquire coefficients. Weβre going to use the latter method. When we distribute the parentheses on our right-hand side, we get two π΄π₯ minus π΄ plus π΅. We begin by equating coefficients of π₯ or π₯ to the power of one. On the left-hand side, thatβs 12. And on the right, itβs two π΄. We divide both sides of this by two. And we find π΄ to be equal to six.

We do this again with π₯ to the power of zero, the coefficients of π₯ to the power of zero. But theyβre just the constant terms. So, on the left-hand side, we have negative four. And on the right, itβs negative π΄ plus π΅. But of course, π΄ is six. So, we find negative four equals negative six plus π΅. And if we add six to both sides, we find π΅ to be equal to two. And so, this means 12π₯ minus four over two π₯ minus one all squared is equal to six over two π₯ minus one plus two over two π₯ minus one squared.

Weβre going to clear some space and now perform the integral. Itβs the integral of four π₯ plus four plus six over two π₯ minus one plus two over two π₯ minus one squared with respect to π₯. Weβll do this term by term. When we integrate four π₯, we add one to the exponent and divide by that value. So, we get four π₯ squared over two which is two π₯ squared. The integral of four is four π₯. And then, we know that when we integrate one over ππ₯ plus π, where π and π are constants, we get one over π times the natural log of ππ₯ plus π. And so, six over two π₯ minus one integrates to six times a half times the natural log of the absolute value of two π₯ minus one, which is three times the natural log of the absolute value of two π₯ minus one.

We do still need to integrate two over two π₯ minus one all squared though. And so, weβre going to perform a substitution. Weβre going to let π’ be equal to two π₯ minus one. This means that dπ’ by dπ₯ is equal to two. Now, whilst dπ’ by dπ₯ isnβt a fraction, we do treat it little like one. And we can say that dπ’ must be equal to two dπ₯. And so, by replacing two dπ₯ with dπ’ and two π₯ minus one with π’, we see that we actually need to integrate one over π’ squared with respect to π’ or the integral of π’ to the power of negative two with respect to π’. Well, thatβs π’ to the power of negative one divided by negative one or simply negative π’ to the power of negative one. Now, weβre going to replace this π’ with two π₯ minus one. And of course, we must add our constant of integration π.

And so, weβve used partial fractions to evaluate our integral. And we get two π₯ squared plus four π₯ plus three times the natural log of the absolute value of two π₯ minus one minus two π₯ minus one to the power of negative one plus some constant of integration π.