### Video Transcript

Determine the integral of π₯ squared minus 16 all over three π₯ squared minus 12π₯ with respect to π₯.

In this question, weβre asked to determine the integral of the quotient of two functions. In fact, this is the quotient of two polynomials, so we call this a rational function. But we donβt know in general how to integrate a rational function. So weβre going to need to use some form of manipulation. And in fact, in this case, thereβs a couple of different options. For example, we could use algebraic division to divide our numerator through by our denominator. This would give us a new integrand. And we could then try and integrate this by using our integral rules.

However, thereβs a method we can try before we do this. We can always try factoring our numerator and our denominator. So letβs try factoring both our numerator and our denominator. Letβs start with the quadratic in our numerator. We can see that both π₯ squared and 16 are squares, so we can recall we can factor this by using difference between squares. We have π squared minus π squared is equal to π minus π times π plus π. So by setting π equal to π₯ and π equal to four, we can factor our numerator to get π₯ minus four multiplied by π₯ plus four. But we still need to factor our denominator. We can see that we can take out a shared factor of three in our denominator and a shared factor of π₯ in our denominator.

So by taking out a factor of three π₯ from our denominator, we need to multiply our first term by π₯ to get three π₯ squared. And we need to multiply our second term by negative four to get negative 12π₯. This gives us a new denominator of three π₯ times π₯ minus four. So we were able to rewrite our integral as the integral of π₯ minus four times π₯ plus four all over three π₯ multiplied by π₯ minus four with respect to π₯. And now we can see something interesting. Both our numerator and denominator share a factor of π₯ minus four. We can cancel this shared factor of π₯ minus four in our numerator and our denominator to simplify our integrand. This gives us the integral of π₯ plus four all over three π₯ with respect to π₯.

And this is still difficult to integrate. However, we can make this a lot easier to integrate if we split this fraction into two. Weβll split our integrand into two, giving us the integral of π₯ over three π₯ plus four over three π₯ with respect to π₯. And now we can simplify this even further. In the first term in our integrand, we can cancel the shared factor of π₯, meaning we now just need to integrate one over three plus four over three π₯ with respect to π₯. And we can in fact do this term by term. First, we can integrate the constant one-third by using the power rule for integration. Or we can just remember that π₯ over three will be an antiderivative for one-third. Either way, we get π₯ over three.

Next, to evaluate the integral of our second term, we need to recall our rules for integrating reciprocal functions. We know for any real constant π, the integral of π over π₯ with respect to π₯ is equal to π times the natural logarithm of the absolute value of π₯ plus a constant integration πΆ. And in our case, the value of the constant π is equal to four over three. So we get that the integral of our second term is four over three times the natural logarithm of the absolute value of π₯. And remember, we need to add a constant of integration πΆ. And this gives us our final answer.

Therefore, we were able to show the integral of π₯ squared minus 16 all over three π₯ squared minus 12π₯ with respect to π₯ is equal to π₯ over three plus four over three times the natural logarithm of the absolute value of π₯ plus a constant of integration πΆ.