Question Video: Finding the Integration of a Rational Function Involving Using the Factorisation of the Difference of Two Squares | Nagwa Question Video: Finding the Integration of a Rational Function Involving Using the Factorisation of the Difference of Two Squares | Nagwa

# Question Video: Finding the Integration of a Rational Function Involving Using the Factorisation of the Difference of Two Squares Mathematics • Third Year of Secondary School

## Join Nagwa Classes

Determine β« (π₯Β² β16)/(3π₯Β² β 12π₯) dπ₯.

03:12

### Video Transcript

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

In this question, weβre 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 donβt know in general how to integrate a rational function. So weβre going to need to use some form of manipulation. And in fact, in this case, thereβs 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, thereβs a method we can try before we do this. We can always try factoring our numerator and our denominator. So letβs try factoring both our numerator and our denominator. Letβs 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. Weβll 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 πΆ. 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 πΆ. 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 πΆ.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions